Calulate the molesules mass of a substance whose 7.0% by mass solution in water freezes at `-0.93^(@)C`. the cryosctopic constant of water is `1.86^(@)C kg mol^(-1)` :
A
`140 g mol ^(-1)`
B
`150.5 g mol ^(-1)`
C
`160 g mol ^(-1)`
D
`155 g mol ^(-1)`
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The correct Answer is:
b
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DILUTE SOLUTION
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When 36.0 g of a solute having the empirical formula CH_(2)O is dissovled in 1.20 kg of water, the solution freezes at -0.93^(@)C . What is the moleculer formula of the solute ? ( K_(f) = 1.86^(@)C kg mol^(-1) )
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What mass of NaCl (molar mass= 58.5 g mol^-1 ) must be dissolved in 65.0 g of water to lower the freezing point by 7.50 ^@C ? The freezing point depression constant, K_f , for water is 1.86 K kg mol^-1 . Assume Van's Hoff factor for NaCl is 1.87.
A solution containing 1.8 g of a compound (empirical formula CH_(2)O ) in 40 g of water is observed to freeze at -0.465^(@) C. The molecules formulea of the compound is ( K_(f) of water =1.86kg K mol^(-1) ):
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