A 5.87 %w/v solution (5.87g per 100.0 mL) of starch in water was found to have...

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

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Expert Solution

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 ( W_B ) = 5.87 g

Average Molar mass of starch (M_B) = ?

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 = iW_B R T / M_B V [ because no. of mole(n) = given mass(W) / Molar mass (M) ]

M_B = iW_B R T / pi V

Now substitute the values of i, W_B , R , T , pi and V in this equation , we have

M_B = 1 x 5.87 g x 0.0821 L atm K^-1mol^-1 x 317.1 K / 0.117 atm x 0.100 L

M_B = 152.8 / 0.0117 g mol^-1

M_B = 13059.829 g/mol

M_B = 1.3059829 x 10⁴ g/mol

M_B = 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

queen_honey_blossom answered 1 year ago

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