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Specific gravity of 98% of H2So4 is 1. 8 gram per CC, 50 ml of the solution is mixed with 1750 ml of pure water molarity of resulting solution?
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Specific gravity of 98% of H2So4 is 1. 8 gram per CC, 50 ml of the sol...
98% of H2SO4
So, 98 gm of H2SO4 in 100 gm of solution 
As 50 ml is the volume, 
mass = 50 x 1.8 g/cc = 90gm
So, H2SO4 in 50 ml on 90 gm is = 98% of 90gm = 88.2 gm
Now, total volume of solution is = 50 +1750 = 1800 ml = 1.8 L
number of moles of H2 SO4 = 88.2/98 =0.9mol
Molarity = 0.9/1.8 = 0.5M
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Specific gravity of 98% of H2So4 is 1. 8 gram per CC, 50 ml of the sol...
Specific Gravity of H2SO4
The specific gravity of a substance is the ratio of its density to the density of a reference substance. In the case of liquids, the reference substance is usually water at a specified temperature. The specific gravity of 98% H2SO4 is given as 1.8 grams per cubic centimeter (g/cc).

Mixing the Solution
To determine the molarity of the resulting solution, we need to first understand the concept of molarity. Molarity is defined as the number of moles of solute per liter of solution.

In this case, we are mixing 50 ml of the 98% H2SO4 solution with 1750 ml of pure water. To calculate the molarity, we need to know the amount of solute (H2SO4) present in the 50 ml of the 98% solution.

Calculating the Amount of Solute
To find the amount of H2SO4 in the 50 ml of the 98% solution, we can use the specific gravity. The specific gravity of 1.8 g/cc means that 1 cc of the solution weighs 1.8 grams. Therefore, the weight of the 50 ml of the solution can be calculated as:

Weight of solution = Volume of solution x Specific gravity
Weight of solution = 50 ml x 1.8 g/cc = 90 grams

Calculating the Amount of H2SO4
To find the amount of H2SO4 present in the solution, we need to consider the concentration of the solution. The 98% concentration means that 98 grams of H2SO4 is present in 100 grams of the solution.

Therefore, the weight of H2SO4 in the 50 ml of the solution can be calculated as:

Weight of H2SO4 = Weight of solution x Concentration of H2SO4
Weight of H2SO4 = 90 grams x (98/100) = 88.2 grams

Calculating the Molarity
Now that we know the amount of H2SO4 in the 50 ml of the solution, we can calculate the molarity. Since we are mixing this solution with 1750 ml of pure water, the final volume of the solution will be 50 ml + 1750 ml = 1800 ml.

To convert the weight of H2SO4 to moles, we need to divide it by the molar mass of H2SO4, which is 98 grams/mol. Therefore, the number of moles of H2SO4 in the solution is:

Number of moles = Weight of H2SO4 / Molar mass
Number of moles = 88.2 grams / 98 grams/mol = 0.9 moles

Finally, we can calculate the molarity by dividing the number of moles by the volume of the solution in liters:

Molarity = Number of moles / Volume of solution (in liters)
Molarity = 0.9 moles / 1.8 liters = 0.5 M

Explanation
In summary, the molarity of the resulting solution obtained by mixing 50 ml of the 98% H2SO4 solution with 1750 ml of pure water is 0.5 M.
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Specific gravity of 98% of H2So4 is 1. 8 gram per CC, 50 ml of the solution is mixed with 1750 ml of pure water molarity of resulting solution?
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Specific gravity of 98% of H2So4 is 1. 8 gram per CC, 50 ml of the solution is mixed with 1750 ml of pure water molarity of resulting solution? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Specific gravity of 98% of H2So4 is 1. 8 gram per CC, 50 ml of the solution is mixed with 1750 ml of pure water molarity of resulting solution? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Specific gravity of 98% of H2So4 is 1. 8 gram per CC, 50 ml of the solution is mixed with 1750 ml of pure water molarity of resulting solution?.
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