The boiling point of solution containing 68.4 g of sucrose `("molar mass" = 342 g mol^(-1))` in 100 g of water is `(K_(b) "for water" = 0.512 K. kg mo

Question

The boiling point of solution containing 68.4 g of sucrose `("molar mass" = 342 g mol^(-1))` in 100 g of water is `(K_(b) "for water" = 0.512 K. kg mo

The boiling point of solution containing 68.4 g of sucrose `("molar mass" = 342 g mol^(-1))` in 100 g of water is `(K_(b) "for water" = 0.512 K. kg mol^(-1))`
A. `100.02^(@)C`
B. `98.98^(@)C`
C. `101.02^(@)C`
D. `100.512^(@)C`

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 - C
`Delta T_(b) = K_(b) xx m = K_(b) xx W_(2)/(M_(2) xx W_(1) (kg))`
`=0.512 xx 68.4/(342 xx 0.1) = 1.024 ^(@) C `
` :. T_(b) = T_(b)^(o) + Delta T_(b) =100 ^(@) C + 1.024^(@)C=101.024^(@)C`

4 views Answered 2023-02-05 15:29