A 1.00 g sample contains only glucose (C_6H_12O_8) and (C_12H_12O_12). When the

Question

A 1.00 g sample contains only glucose (C_6H_12O_8) and (C_12H_12O_12). When the sample is dissolved in water to a total solution volume of 25.0 mL, the osmotic pressure of the solution is 3.78 atm at 298 k. Be sure to show your clean and work on the answer sheet to get partial credit. Work out all practice and work on scrap paper. Assume a 1.00 g/mL density. What is the molarity of the solution? How many moles of glucose are there? How many moles of source are there? What is the mass percent composition of glucose in the sample?

Explanation / Answer

Given:

Weight of sample = 1.00 g
volume = 25.0 mL
pressure = 3.78 atm

Temp = 298 K
Density = 1.00 g/ml
Molar mass of glucose (CHO) : 180.16 g/mol
Molar mass of sucrose (CHO) : 342.30 g/mol

Osmotic pressure is given by:

= (n/V) x RT,

where n/V is the combined number of moles of solutes per liter.
(n/V) = /RT

(n/V)= (3.78 atm)/(0.08206 L-atm/mol•K)(298 K) = 0.1546 mol/L
n = (0.1546 mol/L)(0.025 L) = 0.003864 mol

The molarity of the solution is = No. of combined moles / Volume in litres = 0.1456 mol.L-1

Now,

let n = moles of glucose and n = moles of sucrose.

Then

n = 0.003864 mol = n+ n
Also,

n(180.16 g/mol) = mass of glucose and n(342.30 g/mol) = mass of sucrose

Thus we have 2 equationsà
n(180.16) + n(342.30) = 1.00 g (total mass)
. n+ n = 0.003864

n2 = 0.003864 – n1

Thus on solving we get:

n = 0.001985 mol and n = 0.001879 mol

mass of sucrose = (0.001879 mol)(342.30 g/mol) = 0.6431 g
% sucrose by mass = (0.6431 g)/1.00 g) × 100% = 64.31 %

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