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kramer
3 years ago
12

What is the molar concentration of a sucrose solution prepared by dissolving 350.25 g of sucrose is enough deionized water to yi

eld a final solution volume of 500.00 mL?
Chemistry
1 answer:
vovangra [49]3 years ago
3 0

The  molar concentration of a sucrose solution prepared by dissolving 350.25 g of sucrose is enough deionized water to yield a final solution volume of 500.00 mL is 2.048 M.

We must first obtain the number of moles of sucrose in the solution as follows;

Number of moles = mass/ molar

Mass of sucrose = 350.25 g

Molar mass of sucrose = 342 g/mol

Number of moles of sucrose= 350.25 g / 342 g/mol

= 1.024 moles

Recall that;

Number of moles = concentration × volume

concentration = Number of moles/volume

volume of solution = 500.00 mL or 0.5 L

concentration = 1.024 moles/0.5 L

concentration = 2.048 M

Learn more: brainly.com/question/13385951

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We want Length:

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What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as a buffer made from 475 mL of 0.
Hitman42 [59]

Explanation:

The given data is as follows.

      [HCOOH] = 0.2 M,       [NaOH] = 2.0 M,

         V = 500 ml,   [Benzoic acid] = 0.2 M

First, we will calculate the number of moles of benzoic acid as follows.

   No. of moles of benzoic acid = Molarity × Volume

                         = 2 \times 0.475

                         = 0.095 mol

And, moles of NaOH present in the solution will be as follows.

    No. of moles of NaOH = Molarity × Volume

                          = 2 \times 0.025

                          = 0.05 mol

Hence, the ICE table for the chemical equation will be as follows.

         C_{6}H_{5}COOH + NaOH \rightarrow C_{6}H_{5}COONa + H_{2}O

Initial:        0.095           0.05            0             0

Equlbm:  (0.095 - 0.05)  0            0.05

        pH = pK_{a} + log \frac{Base}{Acid}  

              = 4.2 + log \frac{0.05}{0.045}

              = 4.245

For,  

         HCOOH + NaOH \rightarrow HCOONa + H_{2}O

Initial:       0.2x     2(0.5 - x)               0

Equlbm:   0.2x - 2(0.5 - x)                 0             2(0.5 - x)

As,

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Volume of NaOH = (0.5 - 0.464) L

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And, volume of formic acid is 464 mL.

                 

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