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Solutions

Long Answer Type

131. How many mL of 0.1 M HC1 are required to react completely with 1 g mixture of Na₂CO₃and NaHCO₃ containing equimolar amounts of both?

138 Views

132. A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.

117 Views

133. An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₆O₂) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL^–1, then what shall be the molarity of the solution?

253 Views

134. A sample of drinking water was found to be severely contaminated with chloroform (CHCl₃), supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):
(i) express this in percent by mass
(ii) determine the molality of chloroform in the water sample.

169 Views

Short Answer Type

135. The partial pressure of ethane over a solution containing 6.56 x 10^–3 g of ethane is 1 bar. If the solution contains 5.00 x 10^–2 g of ethane, then what shall be the partial pressure of th gas?

192 Views

136. An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

163 Views

Long Answer Type

137. Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure, in bar of a mixture of 26.0 g of heptane and 35.0 g of octane?

173 Views

138. The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molar solution of a non-volatile solute in it.

216 Views

139. Calculate the mass of a non-volatile solute (molecular mass 40 g mol^–1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

483 Views

Short Answer Type

140. If the density of some lake water is 1.25 g mL^–1 and contains 92 g of Na⁺ ions per kg of water, calculate the molality of Na⁺ ions in the lake.

Answer:

number of moles present in 92g of Na⁺ ion.
=92/23g mol^-1 = 4mol

Molality, $m = \frac{No . of gm moles of solute}{wt . of solvent in kg}$
No. of gm moles of solute = 92/23 = 4
Wt. of water (solvent) = 1 kg
$Molality = \frac{4}{1} = 4 m .$