Question

10 cm3 of human urine moles found to have 5mg of urea calculate the molality of given sample with respect to urea

Answer 1

Answer:

0.00833 mol/kg

Explanation:

Since urine is mostly water, so let's consider the density of water to be 1g/ml.

Mass of the 10 cm3(mL) urine will be 10g.

Considering the the weight of the solvent i.e. water is unaffected by the weight of urea because urea is present in very less amount. So we can take mass of solvent as 10g (even if we want to consider urea we will get 9.995 g of solvent)

Molality = number of moles of solute/ mass of solvent in kg

number of moles of urea= $5*10^{-3}/60$

=$8.33*10^{-5}$

mass of solvent= 0.01 kg

$Molality= \frac{8.33*10^{-5}}{0.01}$

=0.00833 mol/kg

Answer 2

Answer:

0.008333 moles / kg

Explanation:

We first write the chemical formula of urea.

Urea = CH4N2O

Molar mass of urea = 12 + 1 × 4 + 14 × 2 + 16 = 12 + 4 + 28 + 16 = 60 grams/mol

We convert the given mass of urea to grams :

5/1000 = 0.005 grams

Moles of urea = 0.005/60 = 0.00008333 moles = 1/12000 moles

Molality = moles of solute / kilogram of solvent.

The main solvent in urine is water.

Density of water = 1 g/cm³

The mass of solvent = 1 × 10 = 10 grams.

We convert this to kilograms = 10/1000 = 0.01 kilograms.

1/12000 moles —> 0.01 kilograms

? —> 1 kilogram

By cross multiplication we have :

1/0.01 × 1/12000 = 0.008333 moles / kg