Find concentration in mol/dm in 49g in Hydrogen sulphate (vi) in 2dm ?
Question
Find concentration in mol/dm in 49g in Hydrogen sulphate (vi) in 2dm ?
Solution
Sure, here are the steps to find the concentration in mol/dm^3 of 49g of Hydrogen sulphate (vi) in 2dm^3:
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First, we need to find the molar mass of Hydrogen sulphate (vi). The molar mass of Hydrogen is 1g/mol, and the molar mass of Sulphate (vi) is 96g/mol. So, the molar mass of Hydrogen sulphate (vi) is 1 + 96 = 97g/mol.
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Next, we calculate the number of moles of Hydrogen sulphate (vi) in 49g. We use the formula: number of moles = mass / molar mass. So, the number of moles = 49g / 97g/mol = 0.505 moles.
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Finally, we find the concentration in mol/dm^3. We use the formula: concentration = number of moles / volume. The volume is given as 2dm^3. So, the concentration = 0.505 moles / 2dm^3 = 0.2525 mol/dm^3.
So, the concentration of 49g of Hydrogen sulphate (vi) in 2dm^3 is 0.2525 mol/dm^3.
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