## Some Basic concepts in Chemistry

# Mole Concept and Molar Mass

**Mole :** Mole is the seventh base S.I unit which expresses the amount of substance.

\tt Number \ of \ moles = \frac{mass \ of \ substance}{m\left(' g ' \ molecular \ mass\right)}

\tt n_{H_{2}O} = \frac{w}{18} , \tt n_{CH_{4}} = \frac{w}{16}

**Mole Fraction (X) :**

- \tt X_{Solvent} = \frac{moles \ of \ solvent (N)}{moles \ of \ solute (n) \ + \ moles \ of \ solvent(N)}
- \tt X_{Solute} = \frac{moles \ of \ solute (n)}{moles \ of \ solute (n) \ + \ moles \ of \ solvent(N)}
- \tt X_{Solute} = \frac{\frac{w}{m}}{\frac{w}{m} + \frac{W}{M}} \ and \ X_{Solvent} = \frac{\frac{W}{M}}{\frac{w}{m} + \frac{W}{M}}

\tt \therefore X_{solute} + X_{Solvent} = 1

**Molarity (M) :**

- \tt Molarity (M) = \frac{n}{V_{L}} = \ n \times \frac{1000}{V_{ml}} = \frac{w}{m}\times\frac{1000}{V_{ml}}
- \tt M \times V_{ml} = \frac{w}{m} \times 1000 = millimoles
- \tt M \times V_{L} = \frac{w}{m} = moles

**Molality (m) :**

\tt Molality (m) = \frac{moles \ of \ solute (n)}{mass \ of \ solvent (W \ kg)}

**Formality (F) :**

\tt F = \frac{w}{f} \times \frac{1000}{V_{ml}}

f = formula mass of solute

F = Formality

**Normality (N) :**

- No. of equivalents of solute present in 1L of solution.
- \tt N = \frac{E.q's\left\{solute\right\}}{V_{L} of \ solution} \ or \ \frac{no.of \ milli \ equivalents}{V_{mL}}
- \tt N = \frac{Equivalents \times 1000}{V_{ml}}
- No. of milli equivalents of solute = N × V
_{ml}. - \tt N \times V_{L} = \frac{W}{E} = equivalent
- \tt N \times V_{mL} = \frac{W}{E} \times 1000 = m.equivalent

**Relation between concentration units :**

**Molarity and Molality :**

- \tt M = \frac{w}{m^{1}} \times \frac{1000}{V_{ml}} — 1
- \tt m = \frac{w}{m^{1}} \times \frac{1000}{W_{g}} — 2

from 1 and 2 ⇒ \tt w = \frac{M.m^{1} V_{ml}}{1000}

and \tt W = \frac{w \times 1000}{m^{1} \times m}

- \tt density (d) = M\left[\frac{1}{m} + \frac{m^{1}}{1000}\right]

**Molarity and Mole Fraction :**

- \tt M = \frac{X_{solute}}{X_{solvent}}\times \frac{1000 \times W_{g}}{M^{1}_{solvent} \times V_{ml}}
- \tt M_{3} = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}} \ (after \ mixing)

**Molarity and Normality :**

- M × M.wt = N × E.wt
- \tt N = \left(\frac{M.wt}{E.wt}\right) M

N = XM

X = n.factor

### View the Topic in this Video from 0:13 to 7:48

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1. Strength of solution = Amount of substance in g litre^{-1}

2. Strength of solution = Amount of substance in g moles litre^{-1}

3. Strength of solution = Normality × Eq. wt . of the solute in gm/lit

= Molarity × Mol . wt . of solute.

4. \tt Molarity = \frac{Number \ of \ moles \ of \ solute}{Volume \ of \ solution \ in \ litre}

5. \tt Number \ of \ moles = \frac{Wt. \ in \ gm }{Mol. wt.} = M \times V_{\left(in \ l\right)} = \frac{Volume \ in \ litres}{22.4} \ at \ NTP \ \left(only \ for \ gases\right)

6. \tt Number \ of \ millimoles = \frac{Wt.in \ gm \times 1000 }{Mol.wt.} = Molarity \times Volume \ in \ ml.

7. \tt Number \ of \ equivalents = \frac{Wt.in \ gm}{Eq.wt.} = x \times No.of \ moles \times Normality \times Volume \ in \ litre

8. \tt Number \ of \ milliequivalents\left(meq.\right) = \frac{Wt.in \ gm \times 1000}{Eq.wt.} = Normality \times Volume \ in \ mL

9. \tt Normality = x \times No.of \ millimoles = x \times Molarity = \frac{Strength \ of \ solution \ in \ gm \ litre^{-1}}{Eq.wt.of \ solute}

\tt where \ x = \frac{Mol.wt.}{Eq.wt.}, x = Valency \ or \ change \ in \ Oxidation \ Number.

10. \tt Normality \ formula,N_{1}V_{1}=N_{2}V_{2}

11. \tt \% \ by \ weight = \frac{Wt.of \ solute}{Wt.of \ solution}\times100

12. \tt \% \ by \ Volume = \frac{Wt.of \ solute}{Vol.of \ solution}\times100

13. \tt \% \ by \ Strength = \frac{Vol.of \ solute}{Vol.of \ solution}\times100

14. \tt Specific \ gravity = \frac{Wt.of \ solution}{Vol.of \ solution} = Wt.of \ 1 \ mL. of \ solution

15. \tt Formality = \frac{Wt.of \ ionic \ solute}{ Formula \ Wt.of \ solute \times V_{in \ l}}

16. Mol. Wt = V.D × 2 (For gases only)