Question

A nitric acid solution containing 71.0% HNO3 (by mass) is available in a laboratory. How many moles of HNO3 are present in 1.98 L of this solution?
 

a. 

62.8 mol

 

b. 

31.7 mol

 

c. 

22.3 mol

 

d. 

 0.0316 mol

 

e. 

none of these


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Answer 1

The question is incomplete. Here is the complete question.

A nitric acid solution containing 71.0% HNO₃ (by mass) has a density of 1.42g/mL. How many moles of HNO₃ are present in 1.98L of this solution?

a. 62.8 mol

b. 31.7 mol

c. 22.3 mol

d. 3.16 x 10⁻² mol

e. none of these

Answer: b. 31.7 mol

Explanation: The solution has density of 1.42g/mL. So, in 1980 mL of solution:

$d=\frac{m}{V}$

m = d.V

m = 1.42(1980)

m = 2811.6 g

There are 2811.6 grams of solute.

Nitric acid in this solution is 71% by mass, which means

m = 2811.6(0.71)

m = 1996.236 g

In this solution, mass of nitric acid is 1996.236 grams.

Molar mass of HNO₃ is M = 63.01g/mol.

Then:

$n=\frac{m}{M}$

$n=\frac{1996.236}{63.01}$

n = 31.68 moles

In 1.98L of a nitric acid solution, there are 31.7 moles of HNO₃.