Question

What volume of 75% alcohol by weight (d=0.80g/cm3) must be used to prepare 150cm3 of 30% alcohol by weight?

Answer 1

Molarity can be shown as

$M = (W_B /M_B) * (1000 / V)$

$d = W_B /V$

WB is mass in g

V is volume in mL

M_B is molar mass

d is density

$M = (d /M_B) * 1000$

For 70 % solution

$M_1 = (0.8 / M_B) * 1000$

For 30 % solution

$M_2 = (W_B /M_B) * (1000 / V)$

$M_2 = (30 /M_B) * (1000 / 70)$

$M_1 * V_1 = M_2 * V_2$

$[(0.8 / M_B) * 1000 ] * V_1 = (30 /M_B) * (1000 / 70) * 150$

$[800 ] * V_1 = (428.5 ) * 150$

$V_1 = 80.35 mL$

Thus, 80.35 mL of 75 % alcohol is required to dilute upto 150 mL for the formation of 30 % alcoho