Question

1. Apply the like dissoles lihe rule to predict which of the following liquids is immiscible with liquid bromine, B a) chloroform, CHC1, e) water, H,o b) toluene, CJM.CH d) hesane, Call 2, what is the mass of a 1 2.5% blood plasma sample that contains 10.0 g of dissolved solute? a) 1.25 g What is the mass of solute dissolved in 10.0 g of a 5.00% sugar solution? a) 0.180 g b) 1.43 g c)70.0g d) 80.0 g 3, b) 0.500 g e) 0.900 g d) 9.50 g 4. What is the molarity of a glucose solution that contains 10.0 g of CHiz0, (180.18 g/mol) dissolved in 100.0 mL of solution? a) 0.00555 M b) 0.0555 AM c)0.555 M d) 1.80 M 5. What is the mass of nickel(II) nitrate (182.71 g/mol) dissolved in 25.0 mL of 0.100 MNi(NO,)2 solution? a) 0.250g b) 0.457 g c) 4.00 g d) 45.7 g 6. What is the volume of 12.0 Mhydrochloric acid that contains 3.646 g of HCl solute (36.46 g/mol)? a) 1.20 mL b) 8.33 mL c) 83.3 mL d) 120 mL 7. What is the molarity of a nitric acid solution prepared by diluting 250.0 mL of 6.00 ????, to a total volume of 2.50 L? a) 0.0600 M What volume of 16 M nitric acid must be diluted with distilled water to prepare b) 0.250 M c) 0.600 M d) 2.50 M 500.0 mL of 0.50 M HNO3? a) 0.016 mL b) 0.16 mL c) 1.6 mL d) 16 mL What volume of 0.100 M nitric acid reacts completely with 0.500 g of sodiurm carbonate, Na,CO, (105.99 g/mol)? Na2CO(s) + 2 HNO3(aq) ? 2 NaNO3(aq) + H2O(1) + CO2(g) 10.6 m b) 23.6 mL c) 47.2 mL d) 94.3 m

Answer 1

Solution :

1. There are some solutes which are polar and some which are non-polar. The basic principle behind 'like dissolve like' rule is that 'Polar solutes are soluble in polar solvents whereas non polar solutes are soluble in non polar solvents'.

So, we can apply this rule to predict the miscibility of the given solute in the solvents. First, we have to find the polarity of the solvent and all the solute.

Solvent : Liquid Bromine, Br₂ = Non-polar

Solute : Chloroform, CHCl₃ = Non-polar

Toluene, C₆H₅CH₃ = Non-polar

Water, H₂O = Polar

Hexane, C₆H₁₄ = Non-polar

Here, we can see that only Water is polar and all other solutes are non-polar. And it does no match with the with the polarity of liquid bromine which is non-polar. So, by applying the above rule we can predict that water is immiscible with liquid bromine. [Note : This is only a prediction as asked in the question. Experimental result may be different.]

So, the answer is option (c) Water, H₂O.

2. Since, 12.5 % contains 10 g of solute.

The formula is given by ;

There are some solutes which are polar and some which are non-polar. The basic principle behind 'like dissolve like' rule is that 'Polar solutes are soluble in polar solvents whereas non polar solutes are soluble in non polar solvents'. So, we can apply this rule to predict the miscibility of the given solute in the solvents. First, we have to find the polarity of the solvent and all the solute. Solvent: Liquid Bromine, Br2�= Non-polar Solute: Chloroform, CHCl3�= Non-polar Toluene, C6H5CH3�= Non-polar Water, H2O = Polar Hexane, C6H14�= Non-polar Here, we can see that only Water is polar and all other solutes are non-polar. And it does no match with the with the polarity of liquid bromine which is non-polar. So, by applying the above rule we can predict that water is immiscible with liquid bromine. [Note: This is only a prediction as asked in the question. Experimental result may be different.

$or, \frac{12.5}{100} = \frac{10}{Mass\ of\ solution}$

$or, Mass\ of\ solution = \frac{1000}{12.5} = 80.0\ g$

Therefore mass of the solution is 80.0 g. (Option (d)).

3. In sugar solution, solute is sugar and the solvent is water.

5% sugar solution means, 5 g of sugar(solute) dissolved in 100 gm solution.

We can also say, 100 g of solution contains 5 g of sugar.

So, 10 g of solution will contain = ( 5 / 100 ) $\times$ 10 g = 0.5 g ............................ (applying unitary method)

So, the answer is option(b) 0.5 g.

4. The formula for molarity is given by,

$Molarity = \frac{Number\ of\ moles\ of\ solute}{Volume\ of\ solution} = \frac{n}{V}$

Again Number of moles can be given by,

n = ( Given mass / Molar mass ) = ( w / m)

So, molarity is given by ;

$Molarity = \frac{w}{m\times V}$

$or, Molarity = \frac{10}{180.18\times 0.1}$

So, the answer is option (c) 0.555 M.

5. The formula for molarity is given by,

$Molarity = \frac{Number\ of\ moles\ of\ solute}{Volume\ of\ solution} = \frac{n}{V}$

Again Number of moles can be given by,

n = ( Given mass / Molar mass ) = ( w / m)

So, molarity is given by ;

$Molarity = \frac{w}{m\times V}$

$or, 0.100 = \frac{w}{182.71\times 0.025}$

So, the answer is option (b) 0.457 g.

6. The formula for molarity is given by,

$Molarity = \frac{Number\ of\ moles\ of\ solute}{Volume\ of\ solution} = \frac{n}{V}$

Again Number of moles can be given by,

n = ( Given mass / Molar mass ) = ( w / m)

So, molarity is given by ;

$Molarity = \frac{w}{m\times V}$

$or, 12.0 = \frac{3.646}{36.46\times V}$

7. Relationship between molarity and volume is given by ;

$M_{1}\times V_{1} = M_{2}\times V_{2}$

M₁ = Initial molarity of the solution. = 6.00 M

V₁ = Initial volume of the solution. = 250.0 mL

M₂ = Final molarity of the solution. = To be calculated

V₂ = Final volume of the solution. = 2.5 L = 2500 mL

$M_{1}\times V_{1} = M_{2}\times V_{2}$

$or, 6\times 250 = M_{2}\times 2500$

So, the answer is option (c) 0.600 M.