It is the no. of moles of solute present in per litre solution. Unit = Moles per lit.

Molarity Formula is given as

If Volume = 1 litre

M = n solute

Unit of Molarity is Moles L

In terms of weight, molarity formula of the substance can be expressed as:

Molarity formula with Density is given as

Let's take a example of NaOH

Molar mass of NaOH = Molar mass of Na + Molar mass Of O + Molar mass of H

=23+16+1=40

2. We calculate the Moles of the solute by the formula

$n=\frac {\text{Mass(g)}}{\text{Molar Mass}}$

3. Find the Volume of Solution in Litre

We can use below table for our convenience

1 cm^{3} |
1mL |
1000 mm^{3} |

1 Litre |
1000mL |
1000 cm^{3} |

1 m^{3} |
10^{6 }cm^{3} |
1000 L |

1 dm^{3} |
1000 cm^{3} |
1 L |

$M= \frac {\text{No Of Moles of Solute}}{\text {Volume of Solution in (L)}}$

Calculate the molarity of $NaOH$ in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution.

As per Molarity Formula

$M = \frac {W}{M} \times \frac {1000}{\text{V ml}}$

= (4 g / 40 g ) × (1000/250) =0.1 mol/0.250 L

= 0.4 mol/L

= 0.4 M

If 2 gm $NaOH$ is dissolute in water to make solution upto 250 cc. Give molarity of the solution:

NaOH = W = 2 g

Molar Mass of NaOH = 23 + 16 + 1 = 40

M = W/M × 1000/(V ml) = 2/40 × 1000/250=1/5 M

Give the mass of NaOH dissolute to make solution of 500 cm

M = W/M × 1000/(V ml)

Molar Mass of NaOH = 23 + 16 + 1 = 40

M=1/2 and V=500 ml

Therefore,

1/2=W/40 ×1000/500

W=20/2=10gm

20% (w/W) NaOH solution (density of solution= 1.2 g/cc). Calculate the Molarity

20% (w/M) means 20 gm in 100gm of solution

Now Density of solution =1.2g/cc

So Volume of Solution =100/1.2 cc

Now

M = W/M × 1000/(V ml)

M= 20/40 × 1000/100 × 1.2 = 6 M

Give the volume of the solution if 10 gm NaOH, 1/20 M NaOH sol. is present.

M = W/M × 1000/(V ml)

1/20=10/40 × 1000/(V ml)

4/20= 1000/(V ml)

V = 5000 ml

= 5L

What is the amount of solute present in M/40,100 ml Na OH sol.

M = W/M × 1000/(V ml)

1/40=W/40 ×1000/100

4/40 = W

0.1 gm = W

Calculate the morality of 2% (m/v) glucose sol.)

2 gm Glucose in 100 ml sol.

Now Molar mass of glucose(C6 H12 O6) = 12×6+1 ×12+6

= 72 + 12 + 96=180 gm

M = W/M × 1000/(V ml)

M = 2/180 ×1000/100

M = 1/9

M

where M

M

V

V

Suppose we have mix three solution of same solute of Molarity M

Molarity of the resulting solution is given by

$M_R= \frac {M_1V_1 +M_2V_2 + M_3V_3}{V_1 + V_2 + V_3}$

If 500 mL of a 5M solution is diluted to 1500 mL, what will be the molarity of the solution obtained?

(i) 1.5 M

(ii) 1.66 M

(iii) 0.017 M

(iv) 1.59 M

M

500 X 5=1500 M

M

If 4 g of NaOH dissolves in 36 g of H

gravity of solution is 1g mL–1).

Mass of NaOH = 4 g

Number of moles of NaOH =4/40

= 0.1 mol

Mass of H2O = 36 g

Number of moles of H

Mole fraction of water

=Number of moles of H

=2/(2+.1) = 0.95

Mole fraction of NaOH =Number of moles of NaOH/ (No. of moles of NaOH + No. of moles of water)

= .1 /(2+.1) = 0.047

Mass of solution = mass of water + mass of NaOH = 36 g + 4 g = 40 g

Volume of solution = 40 × 1 = 40 mL

(Since specific gravity of solution is = 1 g mL–1)

Molarity of solution

=Number of moles of solute/ Volume of solution in litre

=0.1 mol NaOH /.04 = 2.5 M

Now As per Molarity Formula

For 0.2M , we need 0.2 moles of HCl in 1 Litre Solution

Now Molecular mass of HCl=34.46 g/mol

1 mole of HCl contain $1 \times 34.46$ g HCl (no.of moles= given mass of solute/ molecular mass of solute)

0.2 moles of HCl contain $0.2 \times 34.46=6.892$g HCl

Therefore 6.892 g of HCl in a litre of solution gives 0.2 molar HCl solution

χ

n

n

M = molarity of solution

d = Density of solution

M

M

Now Mass of solution = n

Volume of solution = Mass of solution/density of solution

So, Volume of solution = (n

The density of a solution is in g/mL

Molarity = Number of moles of SOlute/ Volume of the solution

=n

Dividing Numerator and denominator by n

= χ

If density is in g/litre then the molarity is given as

= χ

Check out Molarity Calculator

- Importance Of chemistry
- |
- Physical States of matter
- |
- Chemical Classification Of matter
- |
- The International System of Units (SI units)
- |
- Significant Figures
- |
- Laws of Chemical Combination
- |
- Atomic Mass
- |
- Molecular Mass
- |
- Mole Concept (Avogadro Constant) And Molar mass
- |
- Gram Atomic Mass
- |
- Gram Molecular Mass
- |
- Percentage composition
- |
- StioChiometry
- |
- Mole Fraction
- |
- Molarity Defination
- |
- Molality Defination
- |
- Normality Formula

Class 11 Maths Class 11 Physics Class 11 Chemistry

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