The molal freezing point depression constant =kf6.19·°c·kgmol−1 for a certain substance x . when 10.g of urea nh22co are dissolved in 500.g of x , the solution freezes at −6.8°c . calculate the freezing point of pure x . be sure your answer has the correct number of significant digits.

-4.7°C

Explanation:

The freezing point depression is a colligative property that can be calculated using the following expression.

ΔTf = Kf . b

where

ΔTf: depression in the freezing point

Kf: molal freezing point depression constant

b: molality

The moles of urea (solute) are:

The mass of solvent (X) is 500 g (0.500 kg).

The molality of the urea is:

ΔTf = Kf . b = (6.19 °C.kg/mol) × 0.34 mol/kg = 2.1°C

The freezing point of pure X is:

-6.8°C + 2.1°C = -4.7°C

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g