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Some Basic Concepts of Chemistry question

2025 · 28 Jan · Shift 1 · Q23
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Some Basic Concepts of Chemistry question

2025 · 28 Jan · Shift 1 · Q23

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
The molarity of a 70%70 \%70%(mass/mass) aqueous solution of a monobasic acid (X) is ‾\underline{\hspace{2cm}}​×10−1\times 10^{-1}×10−1 M (Nearest integer) [Given: Density of aqueous solution of (X) is 1.25 g mL−11.25 \mathrm{~g} \mathrm{~mL}^{-1}1.25 g mL−1 Molar mass of the acid is 70 g mol−170 \mathrm{~g} \mathrm{~mol}^{-1}70 g mol−1 ]
Numerical answer
View written solutionFree

Correct answer: 125

  1. Interpret the given data

A 70%70\%70% (m/m)(m/m)(m/m) aqueous solution means:

  • 70 g70\,\text{g}70g of acid is present in 100 g100\,\text{g}100g of solution.

Given:

  • Density of solution =1.25 g mL−1= 1.25\,\text{g mL}^{-1}=1.25g mL−1
  • Molar mass of acid =70 g mol−1= 70\,\text{g mol}^{-1}=70g mol−1

We need the molarity, i.e.

Molarity=moles of solutevolume of solution in litre\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in litre}}Molarity=volume of solution in litremoles of solute​
  1. Take 100 g of solution

In 100 g100\,\text{g}100g solution:

  • Mass of acid =70 g= 70\,\text{g}=70g

So, moles of acid:

moles=7070=1 mol\text{moles} = \frac{70}{70} = 1\,\text{mol}moles=7070​=1mol
  1. Find volume of 100 g solution

Using density:

Volume=massdensity=1001.25=80 mL\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{100}{1.25} = 80\,\text{mL}Volume=densitymass​=1.25100​=80mL

Convert to litres:

80 mL=0.080 L80\,\text{mL} = 0.080\,\text{L}80mL=0.080L
  1. Calculate molarity
M=10.080=12.5 MM = \frac{1}{0.080} = 12.5\,\text{M}M=0.0801​=12.5M

The question asks in the form:

‾×10−1 M\underline{\hspace{2cm}} \times 10^{-1}\,\text{M}​×10−1M

Now,

12.5 M=125×10−1 M12.5\,\text{M} = 125 \times 10^{-1}\,\text{M}12.5M=125×10−1M

So the required nearest integer is:

125\boxed{125}125​
  1. Comparison with stored answer

Stored correct answer = 125125125

This matches our derived answer.

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