Write down the concentration of each of the following solutions in g/L and mol/L.

a. 7.3g HCl in 100ml solution

b. 2g NaOH in 50ml solution

c. 3g CH₃COOH in 100ml solution

d. 4.9g H₂SO₄in 200ml solution

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Solution

a)7.3 g HCl in 100 mL solution:

$Concentration of HCl in g L^{- 1} = \frac{M a s s o f s o l u t e in gram}{Volume of HCl in l i t r e}$

$= \frac{7.3 \times 1000}{100} = 73$ g L^-1

In terms of gram per litre:
7.3 g of HCl in 100 mL has concentration = 73 g L^{-1

Molecular mass of HCl = $1 + 35.5 = 36.5$}

$Molarity = \frac{Mass of solute in m o l e s}{Volume of HCl in L}$

$= \frac{7.3 \times 1000}{36.5 \times 100} = 2$ mol L^-1

In terms of moles per litre:
7.3 g of HCl in 100 mL has concentration = 2 mol L^-1

b)2g NaOH in 50 mL solution

$C o n c e n t r a t i o n o f N a O H i n g L^{- 1} = \frac{M a s s o f N a O H i n g}{V o l u m e o f N a O H i n L}$

$= \frac{2 \times 1000}{50} = 40$ g L^-1
In terms of gram per litre:
2 g of NaOH in 50 mL has concentration = 40 g L^{-1

Molecular mass of NaOH = $23 + 16 + 1 = 40$}

$Molarity = \frac{Mass of solute in m o l e s}{Volume of N a O H in L}$

$= \frac{2 \times 1000}{40 \times 50} = 1$ mol L^-1

In terms of moles per litre:
2 g of NaOH in 50 mL has concentration = 1 mol L^-1

c)3 g CH₃COOH in 50 mL solution

$C o n c e n t r a t i o n o f C H_{3} C O O H i n g L^{- 1} = \frac{M a s s o f C H_{3} C O O H i n g}{V o l u m e o f C H_{3} C O O H i n L}$

$= \frac{3 \times 1000}{100} = 30$ g L^-1
In terms of gram per litre:
3 g of CH₃COOH in 100 mL has concentration = 30 g L^{-1

Molecular mass of} CH₃COOH^{= $2 \times 12 + 2 \times 16 + 4 \times 1 = 60$}

$M o l a r i t y = \frac{M a s s o f s o l u t e i n m o l e s}{V o l u m e o f s o l u t i o n i n L}$

$= \frac{3 \times 1000}{60 \times 100} = 0.5$ mol L^-1

In terms of moles per litre:
3 g of CH₃COOH in 100 mL has concentration = 0.5 mol L^-1

d)4.9 g H₂SO₄ in 200 mL solution

$C o n c e n t r a t i o n o f H_{2} S O_{4} i n g L^{- 1} = \frac{M a s s o f H_{2} S O_{4} i n g}{V o l u m e o f H_{2} S O_{4} i n L}$

$= \frac{4.9 \times 1000}{200} = 24.5$ g L^-1
In terms of gram per litre:
4.9 g of H₂SO₄ in 200 mL has concentration = 24.5 g L^{-1

Molecular mass of} H₂SO₄^{= $2 \times 12 + 2 \times 16 + 4 \times 1 = 60$}

$M o l a r i t y = \frac{M a s s o f s o l u t e i n m o l e s}{V o l u m e o f s o l u t i o n i n L}$

$= \frac{4.9 \times 1000}{98 \times 200} = 1$ mol L^-1

In terms of moles per litre:
4.9 g of H₂SO₄ in 200 mL has concentration = 1 mol L^-1

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