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Molarity Quiz 2

chemistry

 Find the molarity of solution which contains $2 g$ of a chemical in $100 ml$. $\text{(The molecular weight of the chemical is 200 amu).}$

 $Molarity(M)=\dfrac{W}{MW}\times \dfrac{1000}{V(ml)} \\ $ W = Weight of the chemical $ \\ $ MW = Molecular weight of the chemical $ \\ $ V = Volume of the solution in ml.

 $\text{Molarity(M)}=\dfrac{W}{MW}\times \dfrac{1000}{V(ml)} \\  M=\dfrac{2}{200}\times \dfrac{1000}{100} \\  M=\dfrac{2000}{20000} \\ M=0.1M $ 

 In the below image, the beaker contains $5 g$ of sand in $100 ml$ of $0.1 M$ sugar solution. Find the concentration of the solution after the addition of $5 g$ of sand to $100 ml$ of $0.1 M$ solution. $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d4d3e944-4030-4eb2-afc6-ebb56c0d080b6472717524850790770.png"/>

  Sand does not contain any sugar in its composition. 

 Due to the absence of sugar content in the sand, the molarity of the solution is not going to affect. 

 The density of $4 M$ sodium hydroxide is $2.9 g/ml$ in $100 ml$. Find the weight of the sodium hydroxide which is dissolved?

  $Molarity(M)=\dfrac{W}{MW}\times \dfrac{1000}{V(ml)} \\ $ W = Weight of the chemical $ \\ $ MW = Molecular weight of the chemical $ \\ $ V = Volume of the solution in ml.

 $ \text{ Molarity(M)}=\dfrac{W}{MW}\times \dfrac{1000}{V(ml)} \\  W=\dfrac{M\times MW\times V}{1000} \\ \Rightarrow \dfrac{4\times 40\times 100}{1000} \\ \Rightarrow 16g $

  $20 ml$ of sodium hydroxide is treated with $0.1 M$ solution of hydrochloric acid and consumed $20 ml$ of hydrochloric acid. Then find the molarity of the sodium hydroxide solution. 

 $\dfrac{{{M}_{1}}{{V}_{1}}}{{{n}_{1}}}=\dfrac{{{M}_{2}}{{V}_{2}}}{{{n}_{2}}} \\ $ Here ${{n}_{1}}$ = Number of moles of solute-1 $ \\ $ ${{n}_{2}}$= Number of moles of solute-2 $ \\ $ ${{M}_{1}}$ = molarity of solution-1 $ \\ $ ${{M}_{2}}$= Molarity of solution-2 $ \\ $ ${{V}_{1}}$  = volume of the solution-1 $ \\ $ ${{V}_{2}}$= volume of the solution-2}

  We know that one mole of sodium hydroxide \text{ is going to react with one moles  of hydrochloric acid. $ \\ $ Here ${{n}_{1}} = $ Number of moles of solute-1 $= 1$ $ \\ $ ${{n}_{2}} = $ Number of moles of solute-2 $= 1$ $ \\ $  ${{M}_{1}} =$ molarity of solution-1 $ = 0.1$ $ \\ $ ${{M}_{2}}=$ Molarity of solution-2 $= ?$ $ \\ $ ${{V}_{1}} =$ volume of the solution-1 $= 20 ml$ $ \\ $ ${{V}_{2}} =$ volume of the solution-2 $= 20 ml$ $\\ \dfrac{{{M}_{1}}{{V}_{1}}}{{{n}_{1}}}=\dfrac{{{M}_{2}}{{V}_{2}}}{{{n}_{2}}} \\  {{M}_{2}}=\dfrac{{{M}_{1}}{{V}_{1}}}{{{n}_{1}}}\times \dfrac{{{n}_{2}}}{{{V}_{2}}} \\ \Rightarrow \dfrac{0.1\times 20}{1}\times \dfrac{1}{20} \\  {{M}_{2}}=0.1M $

 There are two solutions whose concentrations are $1 M (1000ml)$  and $10M (200 ml)$. If the two solutions are going to mix together then what is the concentration of the resulting solution?

 ${{M}_{(mixture)}}=\dfrac{{{M}_{1}}{{V}_{1}}+{{M}_{2}}{{V}_{2}}}{{{V}_{1}}+{{V}_{2}}} \\ $ ${{M}_{(mixture)}}$ = Molarity or concentration of the mixture $ \\ $ ${{M}_{1}}$ = concentration of the solution-1 $ \\ $ ${{M}_{2}}$ = concentration of the solution-1 $ \\ $ ${{V}_{1}}$ = Volume of the solution-1 $ \\ $ ${{V}_{2}}$ = Volume of the solution-2  

${{M}_{(mixture)}} = \text{Molarity or concentration of the mixture}\\ {{M}_{1}} = \text{concentration of the solution-1} = 1 M\\ {{M}_{2}} = \text{concentration of the solution-1 }= 10 M\\ {{V}_{1}} = \text{Volume of the solution-1 }= 1000 ml\\ {{V}_{2}} = \text{Volume of the solution-2}  = 200 ml\\ {{M}_{(mixture)}}=\dfrac{{{M}_{1}}{{V}_{1}}+{{M}_{2}}{{V}_{2}}}{{{V}_{1}}+{{V}_{2}}} \\ \Rightarrow \dfrac{(1)(1000)+(10)(200)}{100+200} \\ \Rightarrow \dfrac{1000+2000}{300} \\ \Rightarrow 10M $