Question

How many milliliters of a 0.266 M LiNO3 solution are required to make 150.0 mL of 0.075 M LiNO3 solution?

42.3 mL

53.2 mL

18.8 mL

23.6 mL

35.1 mL

Answer 1

The prepared solution have 0.075M of LiNO₃

As we know the molarity is the number of moles of solute present in the 1L of solution.

$molarity=\frac{w}{m}\times \frac{1000}{v}$

Here the prepared solution have

w= mass= to be got

m= molar mass of LiNO₃= 68.94g/mol

v=volume = 150mL

Molarity= 0,075M

By applying the law $molarity=\frac{w}{m}\times \frac{1000}{v}$

$0.075M=\frac{w}{68.94}\times \frac{1000}{150}\\ w= \frac{150\times 68.94\times 0.075}{1000}\\ w= \frac{775.575}{1000}\\ w=0.775g$

Hence the prepared solution has 0.775g of LiNO₃solute.

Now for the preparation of 150mL of 0.075M LiNO₃ we need 0.266M of LiNO₃ that carry 0.775g of LiNO₃. For this, the required volume of the solution is x mL(suppose).

We need to find out the value of x

Here the given data are

volume (v)= x mL

Molarity= 0.266M

Mass(w)= 0.775g

Molar mass of LiNO₃= 68.94g/mol

Now apply the rule $molarity=\frac{w}{m}\times \frac{1000}{v}$

$0.266M=\frac{0.775}{68.94}\times \frac{1000}{x}\\ 0.266M=0.0112\times \frac{1000}{x}\\ 0.266M\times x=0.0112\times1000\\ 0.266M\times x=11.24\\ x=\frac{11.24166}{0.266}\\ x=42.3mL$

Hence the volume is 42.3 mL.

So by seeing the above expression the correct answer is 42.3 mL.