Question

What will be the molarity of solution containing 2.8 gram of k o h per 250 ml of solution? Atomic mass of k is equals to 39

Answer 1

Explanation:

M(KOH)=56 g/ml

w=

100% = 2.8%

c=n/V

n=m/M

ρ=1 g/mL.

Suppose, that m solution =100 g then

mKOH =w∙msolution

/100% = 2.8*100/100% = 2.8 g

nKOH=2.8/56 = 0.05 mol

c=0.05 mol/0.1 L = 0.5 mol/L = M/5

Answer 2

Explanation:

Molecular wt. Of KOH =39+16+1=56

No. Of moles=

$2.8 \div 56 = 0.05$

Molarity =

$moles \: of \: solute \: \div voume \: of \: solution \: in \: litres$

Therefore =0.05/250×10^-3

0.05×10^3/250

50/250

0.2