MHT CET 2024 3rd May Evening Shift | Liquid Solution Question 33 | Chemistry

The molal elevation boiling point constant for water is $0.513^{\circ} \mathrm{C} \mathrm{Kg} \mathrm{mol}^{-1}$. Calculate boiling point of solution if 0.1 mole of sugar is dissolved in 200 g water?

$100.513^{\circ} \mathrm{C}$

$100.256^{\circ} \mathrm{C}$

$100.0513{ }^{\circ} \mathrm{C}$

$100.025^{\circ} \mathrm{C}$

MHT CET 2024 3rd May Evening Shift

MCQ (Single Correct Answer)

-0

What mass of solute (molar mass $58 \mathrm{~g} \mathrm{~mol}^{-1}$) is to be dissolved in $2.5 \mathrm{~dm}^3 \mathrm{~H}_2 \mathrm{O}$ to generate osmotic pressure of 0.245 atm at 300 K ?

$$\left(\mathrm{R}=0.0821 \mathrm{~dm}^3 \text { atm } \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)$$

1.0 gram

0.72 gram

1.44 gram

1.75 gram

MHT CET 2024 3rd May Evening Shift

MCQ (Single Correct Answer)

-0

Which of the following equation correctly represents molar mass of a solute by knowing boiling point elevation?

$\mathrm{M}_2=\frac{1000 \times \Delta \mathrm{T}_{\mathrm{b}} \times \mathrm{W}_2}{\mathrm{~K}_{\mathrm{b}} \times \mathrm{W}_1}$

$\mathrm{M}_2=\frac{1000 \times \mathrm{K}_{\mathrm{b}} \times \mathrm{W}_1}{\Delta \mathrm{~T}_{\mathrm{b}} \times \mathrm{W}_2}$

$\mathrm{M}_2=\frac{1000 \times \Delta \mathrm{T}_{\mathrm{b}} \times \mathrm{W}_1}{\mathrm{~K}_{\mathrm{b}} \times \mathrm{W}_2}$

$\mathrm{M}_2=\frac{1000 \times \mathrm{K}_{\mathrm{b}} \times \mathrm{W}_2}{\Delta \mathrm{~T}_{\mathrm{b}} \times \mathrm{W}_1}$

MHT CET 2024 3rd May Morning Shift

MCQ (Single Correct Answer)

-0

A solution of non volatile solute has boiling point elevation 1.75 K . Calculate molality of solution $\left[\mathrm{K}_{\mathrm{b}}=3.5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right]$

$0.77 \mathrm{~mol} \mathrm{~kg}^{-1}$

$0.69 \mathrm{~mol} \mathrm{~kg}^{-1}$

$0.50 \mathrm{~mol} \mathrm{~kg}^{-1}$

$0.35 \mathrm{~mol} \mathrm{~kg}^{-1}$

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