Question

This question has multiple parts. Work all the parts to get the most points. How would you prepare 1.10 L of a 0.70-M solution of each of the following? a H, SO Dilute from "concentrated" (18 M ) sulfuric acid m L of concentrated H2SO4 pt 1 pt b HCl from concentrated" (12 M) reagent 1 pt Dilute mL of concentrated HCI NICI, from the salt NiCl 64,0 Dissolve g of NiCl . 61,0 in water, and add water until the total volume of the d Sodium carbonate from the pure solid Dissolve g of Na, CO, in water, and add water until the total volume of the Submit Answer Try Another Version 6 item attempts remaining

Answer 1

Answer:-

(a)-

Given:-

molarity of solution (M₁) = 0.70 M

volume of solution (V₁) = 1.10 L = 1.10 $\times$ 1000 = 1100 mL

molarity of concentrated H₂SO₄ (M₂) = 18 M

volume of concentrated H₂SO₄ (V₂) = ?

According to the formula

M₁V₁ = M₂V₂

therefore

molarity of solution (M₁) $\times$ volume of solution (V₁) = molarity of concentrated H₂SO₄ (M₂) $\times$ volume of concentrated H₂SO₄ (V₂)

volume of concentrated H₂SO₄ (V₂) = molarity of solution (M₁) $\times$ volume of solution (V₁) / molarity of concentrated H₂SO₄ (M₂)

volume of concentrated H₂SO₄ (V₂) = 0.70 M $\times$ 1100 mL / 18 M

volume of concentrated H₂SO₄ (V₂) = 770 mL / 18

volume of concentrated H₂SO₄ (V₂) = 42.78 mL

So

volume of water added = volume of solution (V₁) - volume of concentrated H₂SO₄ (V₂)

volume of water added = 1100 mL - 42.78 mL

volume of water added = 1057.22‬ mL

therefore

Dilute 42.78 mL of concentrated H₂SO₄ with 1057.22‬ mL of water.

(b)-

Given:-

molarity of solution (M₁) = 0.70 M

volume of solution (V₁) = 1.10 L = 1.10 $\times$ 1000 = 1100 mL

molarity of concentrated HCl (M₂) = 12 M

volume of concentrated HCl (V₂) = ?

According to the formula

M₁V₁ = M₂V₂

therefore

molarity of solution (M₁) $\times$ volume of solution (V₁) = molarity of concentrated HCl (M₂) $\times$ volume of concentrated HCl (V₂)

volume of concentrated HCl (V₂) = molarity of solution (M₁) $\times$ volume of solution (V₁) / molarity of concentrated HCl (M₂)

volume of concentrated HCl (V₂) = 0.70 M $\times$ 1100 mL / 12 M

volume of concentrated HCl (V₂) = 770 mL / 12

volume of concentrated HCl (V₂) = 64.17 mL

So

volume of water added = volume of solution (V₁) - volume of concentrated HCl (V₂)

volume of water added = 1100 mL - 64.17 mL

volume of water added = 1035.83‬‬ mL

therefore

Dilute 64.17 mL of concentrated HCl with 1035.83‬ mL of water.

(c)-

Given:-

molarity of NiCl₂.6H₂O solution (M₁) = 0.70 M = 0.70 mol / L

volume of NiCl₂.6H₂O solution (V₁) = 1.10 L = 1.10 $\times$ 1000 = 1100 mL

wt.of NiCl₂.6H₂O = ?

As we know that

molar mass of NiCl₂.6H₂O = molar mass of Ni + 2 $\times$ molar mass of Cl + 6 ( 2 $\times$ molar mass of H + molar mass of O)

molar mass of NiCl₂.6H₂O = 58.69 + 2 $\times$ 35.5 + 6 ( 2 $\times$ 1 + 16)

molar mass of NiCl₂.6H₂O = 58.69 + 71.0 + 6 ( 18)

molar mass of NiCl₂.6H₂O = 58.69 + 71.0 + 108

molar mass of NiCl₂.6H₂O =  237.69 g/mol

Also we know that

molarity of compound = wt. of compound (g) / molar mass of compound (g/mol) $\times$ volume of solution (L)

wt. of compound (g) = molarity of compound $\times$ molar mass of compound (g/mol) $\times$ volume of solution (L)

therefore

wt. of NiCl₂.6H₂O = molarity of NiCl₂.6H₂O solution $\times$ molar mass of NiCl₂.6H₂O (g/mol) $\times$ volume of NiCl₂.6H₂O solution (L)

wt. of NiCl₂.6H₂O = 0.70 mol / L $\times$ 237.69 g/mol $\times$ 1.10 L

wt. of NiCl₂.6H₂O = 183.02 g

So

Dissolve 183.02 g of NiCl₂.6H₂O in water , and add water until the total volume of the 1.10 L or 1100 mL.

(d)-

Given:-

molarity of Na₂CO₃ solution (M₁) = 0.70 M = 0.70 mol / L

volume of Na₂CO₃ solution (V₁) = 1.10 L = 1.10 $\times$ 1000 = 1100 mL

wt.of Na₂CO₃ = ?

As we know that

molar mass of Na₂CO₃ = 2 $\times$ molar mass of Na + molar mass of C + 3 $\times$ molar mass of O

molar mass of Na₂CO₃ = 2 $\times$ 23 + 12 + 3 $\times$ 16

molar mass of Na₂CO₃ = 46 + 12 + 48

molar mass of Na₂CO₃ =  106 g/mol

Also we know that

molarity of compound = wt. of compound (g) / molar mass of compound (g/mol) $\times$ volume of solution (L)

wt. of compound (g) = molarity of compound $\times$ molar mass of compound (g/mol) $\times$ volume of solution (L)

therefore

wt. of Na₂CO₃ = molarity of Na₂CO₃ solution $\times$ molar mass of Na₂CO₃ g/mol) $\times$ volume of Na₂CO₃ solution (L)

wt. of Na₂CO₃ = 0.70 mol / L $\times$ 106 g/mol $\times$ 1.10 L

wt. of Na₂CO₃ =  81.62 g

So

Dissolve 81.62 g of Na₂CO₃ in water , and add water until the total volume of the 1.10 L or 1100 mL​​​​​​​.