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OR NNITON2 (ג) What is the molarity of s whltion mads using 45.0 g of NaCl and enough water to give t hiee of whube? (iid) What is the volume ef a \( 0: 2 \mathrm{M} \) AgNO, soluticen sontaining 85 grams of AgNO? (ikin 1hokgale Zore thited ins mh of a 1.6 M splution of Licl to 1.0 L . detemine the new evesentratisa all ibe whition of LCK1

Ask by Chadwick Mills. in Zambia
Mar 16,2025

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- **Molarity of NaCl Solution:** Approximately 0.26 M - **Volume of 0.2 M AgNO₃ Solution:** Approximately 2.5 L - **New Concentration of LiCl Solution:** 1.6 M

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To find the molarity of a solution made with 45.0 g of NaCl in enough water to make 1.0 L, first, calculate the number of moles of NaCl. The molar mass of NaCl is approximately 58.44 g/mol. Dividing 45.0 g by the molar mass gives about 0.770 moles. Since the total volume is 1.0 L, the molarity is 0.770 M. For the second question, the molar mass of AgNO₃ is approximately 169.87 g/mol. To find moles in 85 g, divide 85 g by the molar mass, which equals about 0.500 moles. To find the volume of a 0.2 M solution containing this amount, use the formula Molarity = Moles/Volume. Therefore, Volume = Moles/Molarity, giving you about 2.5 L of the AgNO₃ solution. To determine the new concentration of a solution when diluting from 1.6 M to 1.0 L, you can use the dilution formula C1V1 = C2V2. If you dilute the original solution and add water to reach a new volume of 1.0 L, you will keep the number of moles of solute constant. After performing the calculations, you can find the new concentration of LiCl in the resulting solution.

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