In: Chemistry
A 5.87 %w/v solution (5.87g per 100.0 mL) of starch in water was found to have an osmotic pressure of 88.9 mmHg at 317.1 kelvin. Calculate the average molar mass of the starch used, taking 1.00 atm = 760.0 mmHg. The value of the gas constant, R , is 0.0821 L atm K-1mol-1. Present your answer in 3 signficant figures
Sol:-
Given
Osmotic pressure (pi) = 88.9 mm Hg = 0.117 atm
because given 1atm = 760.0 mmHg
therefore
1mmHg = 1/760.0 atm
88.9 mmHg = 88.9 / 760.0 atm = 0.117 atm
also Temperature (T) = 317.1 K
Gas constant (R) = 0.0821 LatmK-1mol-1
Volume of solution (V) = 100mL = 0.100 L
weight of solute i.e starch ( WB ) = 5.87 g
Average Molar mass of starch (MB) = ?
also vant's hoff factor (i) of starch = 1 ( because starch is a non-electrolyte)
we know
pi = iCRT
pi = inRT / V [ because , Concentration (C) = no. of moles(n) / volume(V) ]
pi = iWB R T / MB V [ because no. of mole(n) = given mass(W) / Molar mass (M) ]
MB = iWB R T / pi V
Now substitute the values of i, WB , R , T , pi and V in this equation , we have
MB = 1 x 5.87 g x 0.0821 L atm K-1mol-1 x 317.1 K / 0.117 atm x 0.100 L
MB = 152.8 / 0.0117 g mol-1
MB = 13059.829 g/mol
MB = 1.3059829 x 104 g/mol
MB = 1.31 x 104 g/mol ( upto three significant figures)
Hence average Molecular mass of starch = 13059.829 g/mol = 1.31 x 104 g/mol