In the determination of hardness of a sample of water, the following results were obtained:
Volume of sample `H_(2)O=100 mL`
volume of `N//50 Na_(2)CO_(3)` added to it `=20 mL`
volume of `N//50 H_(2)SO_(4)` used to back titrate the unreacted `Na_(2)CO_(3)=10 mL`
Calculate the hardness of water in `gL^(-1)`

Explanation

To determine the hardness of a sample of water based on the provided data, we will follow these steps:

### Step 1: Calculate the volume of reacted Na2CO3
The total volume of Na2CO3 added is 20 mL, and the volume of unreacted Na2CO3 (back titrated with H2SO4) is 10 mL. Therefore, the volume of reacted Na2CO3 is:

\[ \text{Volume of reacted Na2CO3} = \text{Total volume added} - \text{Volume of unreacted Na2CO3} \]
\[ \text{Volume of reacted Na2CO3} = 20 \, \text{mL} - 10 \, \text{mL} = 10 \, \text{mL} \]

### Step 2: Calculate the equivalents of Na2CO3
The normality of Na2CO3 is given as \(N/50\). To find the equivalents of Na2CO3 that reacted, we can use the formula:

\[ \text{Equivalents of Na2CO3} = \text{Normality} \times \text{Volume (in L)} \]
\[ \text{Equivalents of Na2CO3} = \frac{1}{50} \times \frac{10 \, \text{mL}}{1000} = \frac{1}{50} \times 0.01 = 0.0002 \, \text{equivalents} \]

### Step 3: Relate the equivalents of Na2CO3 to CaCO3
From the reaction, we know that the equivalents of Na2CO3 are equal to the equivalents of CaCO3 because they react in a 1:1 ratio:

\[ \text{Equivalents of CaCO3} = \text{Equivalents of Na2CO3} = 0.0002 \, \text{equivalents} \]

### Step 4: Calculate the mass of CaCO3
The equivalent mass of CaCO3 is the molar mass divided by the number of equivalents. The molar mass of CaCO3 is approximately 100 g/mol. Therefore, the equivalent mass is:

\[ \text{Equivalent mass of CaCO3} = \frac{100 \, \text{g/mol}}{2} = 50 \, \text{g/equiv} \]

Now, we can calculate the mass of CaCO3:

\[ \text{Mass of CaCO3} = \text{Equivalents} \times \text{Equivalent mass} \]
\[ \text{Mass of CaCO3} = 0.0002 \, \text{equivalents} \times 50 \, \text{g/equiv} = 0.01 \, \text{g} = 10 \, \text{mg} \]

### Step 5: Calculate hardness in g/L
We have determined that 100 mL of water contains 10 mg of CaCO3. To find the hardness in g/L, we convert the mass in 100 mL to mass in 1 L (1000 mL):

\[ \text{Hardness} = \frac{10 \, \text{mg}}{100 \, \text{mL}} \times 1000 \, \text{mL} = 100 \, \text{mg} = 0.1 \, \text{g/L} \]

### Final Answer:
The hardness of the water sample is **0.1 g/L**.

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