1.0 g of a non-electolyte solute (molar mass 250 g `mol^(-1)`) was dissolved in 51.2 g of benzene. If the `K_(f)` for benzene is 5.12 K kg `mol^(-1)`,

Question

1.0 g of a non-electolyte solute (molar mass 250 g `mol^(-1)`) was dissolved in 51.2 g of benzene. If the `K_(f)` for benzene is 5.12 K kg `mol^(-1)`,

1.0 g of a non-electolyte solute (molar mass 250 g `mol^(-1)`) was dissolved in 51.2 g of benzene. If the `K_(f)` for benzene is 5.12 K kg `mol^(-1)`, the freezing point of benzene will be lowerd by :
A. 0.2 K
B. 0.4 K
C. 0.3 K
D. 0.5 K.

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - b
`DeltaT_(f)=(K_(f)xxW_(B))/(M_(B)xxW_(A))`
`=((5.12 kg mol^(-1))xx(1.0g))/((250gmol^(-1))xx(0.0512kg))=0.4 K`

4 views Answered 2023-02-05 15:29