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Answer :
There are 0.00675 moles of Sr(NO3)2 in the 50.0 mL solution of Sr (NO) 0.135 M and the molarity of the KF solution is approximately 0.500 M
1). To calculate the moles of the Sr(NO3)2 solution, we need to use the given volume and molarity.
Given:
Volume of Sr(NO3)2 solution = 50.0 mL = 0.0500 L
Molarity of Sr(NO3)2 solution = 0.135 M
To calculate the moles, we use the formula:
moles = volume (L) × molarity (mol/L)
moles of Sr(NO3)2 = 0.0500 L × 0.135 mol/L
moles of Sr(NO3)2 = 0.00675 mol
Therefore, there are 0.00675 moles of Sr(NO3)2 in the 50.0 mL solution.
2). To determine the molarity of the KF solution, we need to use the stoichiometry of the reaction.
From the balanced equation:
2 moles of KF react with 1 mole of Sr(NO3)2
Given that 27.00 mL of KF solution reacts completely, we need to find the moles of KF in that volume.
Using the molar volume of KF (which is the same as the molar volume of water), we can convert mL to L:
27.00 mL = 0.02700 L
Using the stoichiometry, we can relate the moles of Sr(NO3)2 to the moles of KF:
moles of KF = (moles of Sr(NO3)2) × (2 moles of KF / 1 mole of Sr(NO3)2)
moles of KF = 0.00675 mol × (2/1)
moles of KF = 0.0135 mol
Now, we can calculate the molarity of the KF solution using the moles and volume:
Molarity of KF solution = moles / volume (L)
Molarity of KF solution = 0.0135 mol / 0.02700 L
Molarity of KF solution ≈ 0.500 M
Therefore, the molarity of the KF solution is approximately 0.500 M.
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