The boiling point of an aqueous solution of a non volatile solute is 1

1.	The boiling point of an aqueous solution of a non volatile solute is 100.15 what is the freezing pointquora
Answer» The freezing point is -0.272 degree Celsius is the freezing point. Explanation: For finding the boiling point we will use the van’t Hoff factor which is $T_b = i \times K_b\times m.$ $K_b$ and $K_f$ for water are 0.512 and 1.86 K $molality^-1.$ ‘i’ is given as 1 because it is a non-volatile SOLUTE. Therefore $0.15 = 1 \times 0.512 \times m.$ This gives us the value of m which is equal to 0.2929. Now when we add the volume of the water we get $m \times v -= m' \times 2v.$ Therefore m' is equal to $\frac{0.2929}{2}.$ which gives us a value of 0.1464. Therefore the temperature of freeing is $1 \times$ $K_f \times m'$ which is $1 \times 0.1464 \times1.86$ which is equivalent to 0.272. Therefore to find the $T_f$ we need to apply the formula of $T_f - Tf_o =0- 0.272$ which gives us a value of -0.272 degree Celsius. Learn more about freezing point The freezing point of pure LIQUID is . If the same amount of non-electrolyte non-volatile impurity is added in and ideal solution is formed on cooling at 200 K, only 30%(by mass) liquid is present and at 201 K, 60% (by mass) liquid is present. The value of is: brainly.in/question/9498647 Calculate the freezing point of a solution CONTAINING 18g of glucose(C6H12O6) and 68.4 g of sucrose(C12H22O11) in 200 g of water.The freezing point of water is 273 K and Kf of water is 1.86 K.Kgmol-1 brainly.in/question/3038203

The boiling point of an aqueous solution of a non volatile solute is 100.15 what is the freezing pointquora

Answer»

The freezing point is -0.272 degree Celsius is the freezing point.

Explanation:

For finding the boiling point we will use the van’t Hoff factor which is

$T_b = i \times K_b\times m.$

$K_b$ and $K_f$ for water are 0.512 and 1.86 K $molality^-1.$

‘i’ is given as 1 because it is a non-volatile SOLUTE.

Therefore $0.15 = 1 \times 0.512 \times m.$

This gives us the value of m which is equal to 0.2929.

Now when we add the volume of the water we get $m \times v -= m' \times 2v.$

Therefore m' is equal to $\frac{0.2929}{2}.$ which gives us a value of 0.1464.

Therefore the temperature of freeing is $1 \times$ $K_f \times m'$ which is $1 \times 0.1464 \times1.86$ which is equivalent to 0.272.

Therefore to find the $T_f$ we need to apply the formula of

$T_f - Tf_o =0- 0.272$ which gives us a value of -0.272 degree Celsius.

Learn more about freezing point

The freezing point of pure LIQUID is . If the same amount of non-electrolyte non-volatile impurity is added in and ideal solution is formed on cooling at 200 K, only 30%(by mass) liquid is present and at 201 K, 60% (by mass) liquid is present. The value of is:

brainly.in/question/9498647

Calculate the freezing point of a solution CONTAINING 18g of glucose(C6H12O6) and 68.4 g of sucrose(C12H22O11) in 200 g of water.The freezing point of water is 273 K and Kf of water is 1.86 K.Kgmol-1

brainly.in/question/3038203