Question

Part A) If you had only 35 g of KOH remaining in a bottle, how many milliliters of 10.0 %(w/v) solution could you prepare?

Express your answer using two significant figures.

Part B)

How many milliliters of 0.27 M solution could you prepare?

Express your answer using two significant figures.

Thanks in advance!

Answer 1

A)

use:

(w/v)% = mass of KOH * 100 / volume of solution

10.0 = 35 g * 100 / volume of solution

volume of solution = 350 mL

Answer: 350 mL

B)

Molar mass of KOH,

MM = 1*MM(K) + 1*MM(O) + 1*MM(H)

= 1*39.1 + 1*16.0 + 1*1.008

= 56.108 g/mol

mass(KOH)= 35 g

use:

number of mol of KOH,

n = mass of KOH/molar mass of KOH

=(35 g)/(56.11 g/mol)

= 0.6238 mol

use:

M = number of mol / volume in L

0.27 = 0.6238/ volume in L

volume = 2.31 L

volume = 2.31*10^3 mL

Answer: 2.3*10^3 mL