What is the molarity of solution containing 10% by mass of calcium hydroxide in 0.5 dm3 of solution?
Answers
What is molarity of 10% NaOH (w/v) solution?
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Antonio Suazo
Answered 11 months ago
The morality is 2.78, must take into account the actual weight of 10 g of NaOH in 100 ml.
M=mol/L NaOH density is 2.13 g/mL
10g Naoh/40(g/mol)=0.25 mol
10% in 100 ml means 10 ml of NaOH and 90 ml of dilutent (water)
2.13*10 (%ofNaOH)= 21.3 is the weight of NaOH
For water grams are equal to mL so 90g water = 90 mL of water
21.3+90=111.3 total weight 111.3g/100 mL= 1.113g/ml density of 10%
Which means 100g/1.113(g/ml)=89.9 ml or 0.0898 L
0.25 mol/0.0898 L=2.78
Answer:
\mathbb{GIVEN}GIVEN
mass percent = 10%
Volume= 0.5 dm³ = 0.5 L
Molar mass of Ca(OH)₂ = 40 + 32 +2 =74 g/mol
\mathbb{TO \: FIND}TOFIND
Molarity
\mathbb{SOLUTION}SOLUTION
Solution containing 10% by mass of calcium hydroxide means 10 g of calcium hydroxide is present in 100 g of solution.
Mass of Ca(OH)₂ = 10g
Molar mass of Ca(OH)₂ = 74 g/mol
\begin{gathered}\begin{gathered}\to\tt{Moles \: of \: Ca(OH)_2= \dfrac{Mass \: of \: Ca(OH) _2}{molar mass \: of \: Ca(OH) _2}} \\\end{gathered}\end{gathered}
→MolesofCa(OH)
2
=
molarmassofCa(OH)
2
MassofCa(OH)
2
\begin{gathered}\begin{gathered}\to\tt{Moles \: of \: Ca(OH)_2= \dfrac{10}{74}} \\\end{gathered} \end{gathered}
→MolesofCa(OH)
2
=
74
10
\begin{gathered}\begin{gathered}\to\tt{Moles \: of \: Ca(OH)_2= 0.135} \\\end{gathered}\end{gathered}
→MolesofCa(OH)
2
=0.135
Now, as per the formula,
\begin{gathered}\begin{gathered}\to\tt{Molarity= \dfrac{moles \: of \: Ca(OH) _2}{Volume \: of \: solution (L) }} \\\end{gathered}\end{gathered}
→Molarity=
Volumeofsolution(L)
molesofCa(OH)
2
\begin{gathered}\begin{gathered}\to\tt{Molarity= \dfrac{0.135}{0.5}} \\\end{gathered}\end{gathered}
→Molarity=
0.5
0.135
\begin{gathered}\begin{gathered}\to {\bf{Molarity= 0.27 \: molar}} \\\end{gathered}\end{gathered}
→Molarity=0.27molar
\fbox{☛Molariry = {0.27} Molar }
☛Molariry = 0.27 Molar