To solve the problem step by step, we will follow the given information and perform the necessary calculations.
### Step 1: Calculate the equivalents of HCl used
We know that:
- 100 mL of 0.5 N NaOH is used to neutralize the unused HCl.
First, we calculate the equivalents of NaOH:
\[
\text{Equivalents of NaOH} = \text{Normality} \times \text{Volume (in L)} = 0.5 \, \text{N} \times 0.1 \, \text{L} = 0.05 \, \text{equivalents}
\]
Since NaOH and HCl react in a 1:1 ratio, the equivalents of HCl that reacted with NaOH is also 0.05 equivalents.
### Step 2: Calculate the total equivalents of HCl initially present
The total volume of HCl used is 150 mL, which is 0.15 L. The normality of HCl is 1 N, so the total equivalents of HCl initially present is:
\[
\text{Total equivalents of HCl} = \text{Normality} \times \text{Volume (in L)} = 1 \, \text{N} \times 0.15 \, \text{L} = 0.15 \, \text{equivalents}
\]
### Step 3: Calculate the equivalents of HCl that reacted with M2CO3
The equivalents of HCl that reacted with M2CO3 can be calculated by subtracting the equivalents of HCl that reacted with NaOH from the total equivalents:
\[
\text{Equivalents of HCl that reacted with M2CO3} = 0.15 - 0.05 = 0.10 \, \text{equivalents}
\]
### Step 4: Relate the equivalents of HCl to M2CO3
Since M2CO3 reacts with HCl in a 1:2 ratio (1 equivalent of M2CO3 reacts with 2 equivalents of HCl), the equivalents of M2CO3 will be half of the equivalents of HCl that reacted:
\[
\text{Equivalents of M2CO3} = \frac{0.10}{2} = 0.05 \, \text{equivalents}
\]
### Step 5: Calculate the molar mass of M2CO3
The molar mass of M2CO3 can be expressed in terms of the equivalent mass of M:
\[
\text{Molar mass of M2CO3} = 2M + 12 + 48 = 2M + 60
\]
Since we have 0.05 equivalents of M2CO3 and we know that:
\[
\text{Equivalents} = \frac{\text{mass}}{\text{equivalent mass}}
\]
We can write:
\[
0.05 = \frac{5.3 \, \text{g}}{\text{Equivalent mass of M2CO3}}
\]
### Step 6: Calculate the equivalent mass of M2CO3
From the above equation, we can find the equivalent mass of M2CO3:
\[
\text{Equivalent mass of M2CO3} = \frac{5.3 \, \text{g}}{0.05} = 106 \, \text{g/equiv}
\]
### Step 7: Relate the equivalent mass of M2CO3 to M
We know that the equivalent mass of M2CO3 is given by:
\[
\text{Equivalent mass of M2CO3} = \frac{2M + 60}{2}
\]
Setting this equal to the calculated equivalent mass:
\[
\frac{2M + 60}{2} = 106
\]
Multiplying both sides by 2:
\[
2M + 60 = 212
\]
Subtracting 60 from both sides:
\[
2M = 152
\]
Dividing by 2:
\[
M = 76
\]
### Step 8: Calculate the equivalent mass of M
The equivalent mass of M is given by:
\[
\text{Equivalent mass of M} = \frac{M}{n}
\]
Where \(n\) is the number of equivalents per mole. For M2CO3, \(n = 2\):
\[
\text{Equivalent mass of M} = \frac{76}{2} = 38
\]
### Final Answer
The equivalent mass of M is 38 g/equiv.
