0-15 g of a substance dissolved in 15 g of a solvent is boiled at a temperature higher by 0-216- C than that of the pure solvent- Find out the molecular weight of the substance Kf for solvent is 2-16 K kg m-1

Amount of solute (w_A)= 0.15 g Amount of solvent (w_B)= 15 g The change in temperature is 0.216^∘ C or 0.216 K. nbsp; K_f= 2.16 K kg m^ 1 The change in te ...

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Standard X

Chemistry

Atomic Mass

0.15 g of a s...

Question

$0.15 g$ of a substance dissolved in $15 g$ of a solvent is boiled at a temperature higher by ${0.216}^{\circ} C$ than that of the pure solvent. Find out the molecular weight of the substance ( $K_{f}$ for solvent is $2.16 {K kg m}^{- 1}$ )

1.01

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10.1

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100

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1000

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Solution

The correct option is C $100$
Amount of solute $(w_{A}) = 0.15 g$
Amount of solvent $(w_{B}) = 15 g$
The change in temperature is ${0.216}^{\circ} C$ or $0.216 K$ .
$K_{f} = 2.16 {K kg m}^{- 1}$
The change in temperature is $Δ T = \frac{1000 \times K_{f} \times w_{A}}{w_{B} \times m_{A}} \Rightarrow 0.216 = \frac{1000 \times 2.16 \times 0.15}{15 \times m_{A}} \Rightarrow m_{A} = 100$

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0.15 g of a substance dissolved in 15 g of a solvent is boiled at a temperature higher by 0.216∘ C than that of the pure solvent. Find out the molecular weight of the substance (Kf for solvent is 2.16 K kg m−1)

The correct option is C 100Amount of solute (wA)=0.15 g Amount of solvent (wB)=15 g The change in temperature is 0.216∘ C or 0.216 K. Kf=2.16 K kg m−1 The change in temperature is ΔT=1000×Kf×wAwB×mA⇒0.216=1000×2.16×0.1515×mA⇒mA=100

$0.15 g$ of a substance dissolved in $15 g$ of a solvent is boiled at a temperature higher by ${0.216}^{\circ} C$ than that of the pure solvent. Find out the molecular weight of the substance ( $K_{f}$ for solvent is $2.16 {K kg m}^{- 1}$ )