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Exercises · 1.6

Q.Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL−11.41\ g\ mL^{-1} and the mass per cent of nitric acid in it being 69%.

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Convert mass percent and density into molarity by finding the mass of acid per litre of solution, then dividing by molar mass. The concentration is 15.44 mol L−1\boxed{15.44\ \text{mol L}^{-1}}.

Why this approach works

Molarity asks "how many moles of solute per litre of solution?" We're given two pieces of information that together let us answer this: the density tells us how much the solution weighs per unit volume, and the mass percent tells us what fraction of that weight is nitric acid. Multiply them to get grams of acid per litre, then convert grams to moles using the molar mass.

The key insight is that density bridges the gap between mass-based composition (mass percent) and volume-based concentration (molarity).


Step-by-step calculation

1. Find the mass of 1 litre of solution

The density is 1.41 g mL−11.41\ \text{g mL}^{-1}. Since 1 L=1000 mL1\ \text{L} = 1000\ \text{mL}:

Mass of 1 L solution=1.41 g mL−1×1000 mL=1410 g\text{Mass of 1 L solution} = 1.41\ \text{g mL}^{-1} \times 1000\ \text{mL} = 1410\ \text{g}

2. Calculate the mass of nitric acid in that litre

The solution is 69% nitric acid by mass, meaning 69 g of HNO3\text{HNO}_3 in every 100 g of solution:

Mass of HNO3=69100×1410 g=973.9 g\text{Mass of HNO}_3 = \frac{69}{100} \times 1410\ \text{g} = 973.9\ \text{g}

3. Determine the molar mass of nitric acid

M(HNO3)=1+14+3(16)=1+14+48=63 g mol−1M(\text{HNO}_3) = 1 + 14 + 3(16) = 1 + 14 + 48 = 63\ \text{g mol}^{-1}

4. Convert mass to moles

n=massmolar mass=973.9 g63 g mol−1=15.46 moln = \frac{\text{mass}}{\text{molar mass}} = \frac{973.9\ \text{g}}{63\ \text{g mol}^{-1}} = 15.46\ \text{mol}

5. Express as molarity

Since we calculated moles in exactly 1 litre of solution:

Molarity=15.46 mol1 L=15.46 mol L−1\text{Molarity} = \frac{15.46\ \text{mol}}{1\ \text{L}} = 15.46\ \text{mol L}^{-1} …

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