h2so4 is labelled as 9.8 by weight. specific gravity of h2so4 is 1.8. the volume of acid to be taken to prepare 100 ml of 0.18M solution is?
M=10×d×y÷gram molecular weight
M=10(1.8)9.8 ÷ 98
M=1.8
M1×V2=M2×V1
(1.8)V2=1000(0.18)
V2=100ml
To determine the volume of H2SO4 needed to prepare a 0.18M solution, we first need to understand the relationship between molarity, weight, and specific gravity.
1. Molarity (M) represents the number of moles of solute dissolved in one liter of solution.
2. Weight percent (w/w%) is the mass of the solute (H2SO4) divided by the mass of the solution, multiplied by 100.
3. Specific gravity (SG) is the ratio of the density of a substance to the density of a reference substance (usually water).
In this case, the concentration of H2SO4 is given as 9.8 by weight, which means 9.8 grams of H2SO4 is present in 100 grams of the solution.
To calculate the molarity (M), we need to convert the weight percent to grams per liter (g/L). Let's assume we have a solution of 100 mL (0.100 L) in volume.
1. Calculate the actual weight of H2SO4 in the solution:
9.8% by weight = (9.8/100) * 100 grams = 9.8 grams
2. Determine the density of H2SO4 using the specific gravity:
Density of water = 1 g/mL
Density of H2SO4 = 1.8 * Density of water = 1.8 g/mL
3. Calculate the volume of H2SO4 required to obtain the desired concentration:
Molarity (M) = Moles of solute / Volume of solution (in L)
Moles of solute = Concentration (M) * Volume of solution (in L)
Rearranging the formula:
Volume of solution (in L) = Moles of solute / Concentration (M)
Moles of solute = (0.18 M) * (0.100 L) = 0.018 moles
Volume of H2SO4 = 0.018 moles / (9.8 g / mol) = 0.018 moles / 9.8 g/mol = 0.00184 L (or 1.84 mL)
Therefore, you would need to take 1.84 mL (or 1.84 cc) of H2SO4 to prepare 100 mL of a 0.18M solution.
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Is that 9.8% w/w? I assume so. Then the molarity of the H2SO4 is
1000 mL x 1.8 g/mL x (9.8/100) x 1/98 = ? which I will label as y.
Then mL1 x M1 = mL2 x M2
....100 mL x 0.18 = mL2 x y
Solve for mL2