MHT CET · Chemistry · Solutions
The solution containing \(6 \mathrm{~g}\) urea (molar mass 60) per \(\mathrm{dm}^3\) of water and another solution containing \(9 \mathrm{~g}\) of solute \(\mathrm{A}\) per \(\mathrm{dm}^3\) water freezes at same temperature. What is molar mass of A?
- A 90
- B 180
- C 54
- D 120
Answer & Solution
Correct Answer
(A) 90
Step-by-step Solution
Detailed explanation
\(\left(\Delta \mathrm{T}_{\mathrm{i}}\right)_1=\left(\Delta \mathrm{T}_{\mathrm{f}}\right)_2 \)
\( \mathrm{~m}_1=\mathrm{m}_2 \)
\( \mathrm{~m}=\text { molality, } 1 \mathrm{dm}^3=1 \mathrm{~L}=1000 \mathrm{ml} \mathrm{H}_2 \mathrm{O} \)
\( 1000 \mathrm{ml} \mathrm{H}_2 \mathrm{O}=1000 \mathrm{~g} \mathrm{H}_2 \mathrm{O} \)
\( \mathrm{m}=\frac{\mathrm{mol} / \mathrm{M}}{\text { Mass of } \mathrm{H}_2 \mathrm{O}} \)
\( \frac{6 / 60}{1}=\frac{9 / \mathrm{M}_{\mathrm{A}}}{1} \)
\( \mathrm{M}_{\mathrm{A}}=\frac{60 \times 9}{6}=90\)
\( \mathrm{~m}_1=\mathrm{m}_2 \)
\( \mathrm{~m}=\text { molality, } 1 \mathrm{dm}^3=1 \mathrm{~L}=1000 \mathrm{ml} \mathrm{H}_2 \mathrm{O} \)
\( 1000 \mathrm{ml} \mathrm{H}_2 \mathrm{O}=1000 \mathrm{~g} \mathrm{H}_2 \mathrm{O} \)
\( \mathrm{m}=\frac{\mathrm{mol} / \mathrm{M}}{\text { Mass of } \mathrm{H}_2 \mathrm{O}} \)
\( \frac{6 / 60}{1}=\frac{9 / \mathrm{M}_{\mathrm{A}}}{1} \)
\( \mathrm{M}_{\mathrm{A}}=\frac{60 \times 9}{6}=90\)
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