Home
Class 12
CHEMISTRY
What mass of NaCl (molar mass = 58.5" g ...

What mass of NaCl (molar mass = `58.5" g "mol^(-1)` be dissolved in 65 g of water to lower the freezing point by `7.5^(@) C`? The freezing pont depression constant, `K_(f)`, for water is 1.86 K kg `mol^(-1)` . Assume van't Hoff factor for Nacl is 1.87.

Text Solution

Verified by Experts

For electrolytes,
`DeltaT_(f)= i xx K_(f) xx m = (i xx K_(f)xx W_(B) xx 1000)/(W_(A) xx M_(B))`
Here, `i=1.87 , K_(f) = 1.86 "K kg mol"^(-1)`,
`M_(B) = 58.5" g mol"^(-1), DeltaT_(f) = 7.5^(@)C or 7.5K`
`W_(A) = 65g, W_(B) = ?`
`7.5=(1.87 xx 1.86 xx W_(B) xx 100)/(65 xx 58.5)`
or `W_(B) = (7.5 xx 65 xx 58.5)/(1.87 xx 1.86 xx 1000) = 8.2 g`
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    ICSE|Exercise Follow up Problems |58 Videos

  • SOLUTIONS

    ICSE|Exercise EXERCISE (PART-I Objective Questions)|26 Videos

  • SOLID STATE

    ICSE|Exercise ISC EXAMINATION QUESTIONS PART-I (Numerical Problems)|6 Videos

  • SOME IMPORTANT ORGANIC NAME REACTIONS

    ICSE|Exercise Questions |278 Videos

Similar Questions

Explore conceptually related problems

What mass of NaCl ("molar mass" =58.5g mol^(-1)) be dissolved in 65g of water to lower the freezing point by 7.5^(@)C ? The freezing point depression constant K_(f) , for water is 1.86 K kg mol^(-1) . Assume van't Hoff factor for NaCl is 1.87 .

What mass of NaCI ("molar mass" =58.5g mol^(-1)) be dissolved in 65g of water to tower the freezing point by 7.5^(@)C ? The freezing point depression constant, K_(f) , for water is 1.86 K kg mol^(-1) . Assume van't Hoff factor for NaCI is 1.87 .

5 g of a substance when dissolved in 50 g water lowers the freezing by 1.2^(@)C . Calculate molecular wt. of the substance if molal depression constant of water is 1.86 K kg mol^-1 .

What mass of ethlene ethyelne glycol ("molar mass"=62 g mol^(-1)) must be dissolved in 5.5kg of water to lower the freezing point of from 0^(@)C to-10^(@)C ? (K_(f)) for water=1.86 K kg mol^(-1) ).

Calculatate the mass of compound (molar mass = 256 g mol^(-1) be the dissolved in 75 g of benzene to lower its freezing point by 0 .48 k(k_(f) = 5. 12 k kg mol ^(-1) .

1.0 g of non-electrolyte solute dissolved in 50.0 g of benzene lowered the freezing point of benzene by 0.40 K . The freezing point depression constant of benzene is 5.12 kg mol^(-1) . Find the molecular mass of the solute.

6 g of a substance is dissolved in 100 g of water depresses the freezing point by 0.93^(@)C . The molecular mass of the substance will be: ( K_(f) for water = 1.86^(@)C /molal)

How much amount of NaCl should be added to 600 g fo water (varphi =1.00 g//mL) to decrease freezing point of water to -0.2^(@)C ? ___________________. (The freezing point depression constant for water =2K kg "mol"^(-1) )

1.00 g of a non-electrolyte solute (molar mass 250g mol^(–1) ) was dissolved in 51.2 g of benzene. If the freezing point depression constant K_(f) of benzene is 5.12 K kg mol^(–1) , the freezing point of benzene will be lowered by:-

1.00 g of a non-electrolyte solute (molar mass 250g mol^(–1) ) was dissolved in 51.2 g of benzene. If the freezing point depression constant K_(f) of benzene is 5.12 K kg mol^(–1) , the freezing point of benzene will be lowered by:-