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10 millimoles of solute is present in th...

10 millimoles of solute is present in the following volume of 0.08M solution

A

125 ml

B

625 ml

C

500 ml

D

1000 ml

Text Solution

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The correct Answer is:
To solve the problem of finding the volume of a solution when given the amount of solute and its molarity, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Given Information**: - We have 10 millimoles of solute. - The molarity (M) of the solution is 0.08 M. 2. **Convert Millimoles to Moles**: - Since 1 millimole is equal to \(10^{-3}\) moles, we can convert 10 millimoles to moles: \[ \text{Number of moles} = 10 \, \text{mmol} = 10 \times 10^{-3} \, \text{moles} = 0.01 \, \text{moles} \] 3. **Use the Molarity Formula**: - The formula for molarity (M) is given by: \[ M = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \] - Rearranging this formula to find the volume of the solution gives us: \[ \text{Volume of solution} = \frac{\text{Number of moles of solute}}{M} \] 4. **Substitute the Values**: - Now, substituting the values we have: \[ \text{Volume of solution} = \frac{0.01 \, \text{moles}}{0.08 \, \text{M}} = 0.125 \, \text{liters} \] 5. **Convert Liters to Milliliters**: - Since the answer needs to be in milliliters, we convert liters to milliliters using the conversion factor \(1 \, \text{liter} = 1000 \, \text{milliliters}\): \[ 0.125 \, \text{liters} = 0.125 \times 1000 \, \text{milliliters} = 125 \, \text{milliliters} \] 6. **Final Answer**: - Therefore, the volume of the solution is **125 milliliters**.

To solve the problem of finding the volume of a solution when given the amount of solute and its molarity, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Given Information**: - We have 10 millimoles of solute. - The molarity (M) of the solution is 0.08 M. ...
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Knowledge Check

  • Number of milli equivalents of solute present in 250 ml of 0.1 M oxalic acid solution are

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    25
    B
    50
    C
    250
    D
    125
  • The number of iodine atoms present in 40mL solution of its 0.1M solution is

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    `4.81 xx 10^(21)`
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    `24.08 xx 10^(21)`
    C
    `0.408 xx 10^(23)`
    D
    `6.02 xx 10^(22)`
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    B
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    D
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