Question

what will be the molarity of the solution obtained by dissolving 4.9g h2so4 in 250g of water​

Answer 1

Answer:

Molar mass of sulphuric acid is 98.07g/mol

Therfore 4.9g contains 0.04996 moles.

0.04996 mol/500ml =( 0.04996 x2) moles per litre = 0.09992 moles per litre

Therefore 4.9g H2SO4 in 500 ml = approx. 0.10M

Answer 2

Given:

Mass of $H_{2}SO_{4}$ = 4.9 g

Volume of water = 250 g

To find:

Molarity of the solution = ?

Formula to be used:

No. of moles = Mass / Molar mass

Molarity, (M) = No. of moles / Volume

Calculation:

Step 1 of 2

Find number of moles:

To find the number of moles, we need the molar mass of  $H_{2}SO_{4}$.

The standard molar mass value of  $H_{2}SO_{4}$ = 98 g

Therefore,

No. of moles = Mass / Molar mass

No. of moles = $\frac{4.9}{49}$

No. of moles = 0.1

Step 2 of 2

Find the molarity by using the given and obtained values:

Molarity, (M) = No. of moles / Volume

Molarity = $\frac{0.1}{250}$

Molarity = 0.0004

Final answer:

The molarity of the solution obtained by dissolving 4.9 g of  $H_{2}SO_{4}$ in 250 g of water is 0.0004 M.

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