How many kg of wet NaOH containing 15% water are required to prepare 70 litre of 0.5N solution?
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0.3
To solve this problem, we first need to determine the amount of actual NaOH in the wet mixture. Then we can use this information to calculate the number of kilograms needed to prepare the desired solution.
Step 1: Calculate the amount of NaOH in the wet mixture.
The wet mixture contains 15% water, which means the remaining 85% is NaOH. We can express this as a fraction: 85/100 = 0.85. So, the wet mixture contains 0.85 grams of NaOH for every gram of the wet mixture.
Step 2: Convert the volume of the desired solution from liters to milliliters.
There are 1000 milliliters in one liter. Therefore, 70 liters = 70 x 1000 = 70,000 milliliters.
Step 3: Calculate the number of moles of NaOH needed for the desired solution.
We want to prepare a 0.5N (normal) solution of NaOH. This means that for every 1 liter (1000 milliliters) of solution, there should be 0.5 moles of NaOH. Since we have 70,000 milliliters of solution, we can calculate the number of moles needed as follows:
0.5 moles/L x 70 L = 35 moles.
Step 4: Convert moles to grams.
The molar mass of NaOH is 22.99 g/mol for Na, 16 g/mol for O, and 1.01 g/mol for H. Adding these together gives a molar mass of 39.99 g/mol for NaOH.
To convert moles to grams, we can multiply the number of moles by the molar mass:
35 moles x 39.99 g/mol = 1,399.65 grams.
Step 5: Convert grams to kilograms.
Since the question asks for the amount in kilograms, we need to divide 1,399.65 grams by 1000 to get the value in kilograms:
1,399.65 grams / 1000 = 1.39965 kilograms.
Therefore, approximately 1.4 kg of wet NaOH containing 15% water is required to prepare 70 liters of a 0.5N solution.