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shusha [124]
2 years ago
14

A student prepares a standardized solution of sodium hydroxide by the procedure outlined. The student first prepares a saturated

solution of sodium hydroxide using freshly distilled or deionized water. She measures about 5 mL of the saturated solution with a graduated cylinder and dilutes to 1 L in a plastic bottle with freshly distilled or deionized water to make a solution that is about 0.1 M. To determine the exact concentration, she puts the NaOH solution in a buret and titrates a carefully-weighed pre-dried solid acid of known molar mass. She records the volume of NaOH needed to neutralize it and repeats the process several times to determine the concentration of the NaOH solution. She calculates the concentration and uncertainty in the concentration of the NaOH from the titration data.
Chemistry
1 answer:
skad [1K]2 years ago
8 0

NaOH has the property of deliquescent hence it cannot be weighed directly instead a saturated solution is diluted and desired concentration is obtained.

Explanation:

Sodium hydroxide undergoes deliquescent.

Deliquescence is the process in which substance absorbs moisture and carbon dioxide from the atmosphere to the extent of getting dissolved and form a solution.

Deliquescence happens when vapour pressure of solution formed is less than the partial pressure of water vapour in the air.

This is the reason that student did not measure the sodium hydroxide directly as the weight will change and proper solution cannot be made. Because of the deliquescence standard solution of NaOH is not prepared.

The student made a solution without weighing the NaOH instead she prepared a solution till the NaOH got dissolved and then diluted the solution.

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Tamiku [17]

Explanation:

It is known that relation between entropy, heat energy and temperature is as follows.

        dS = \frac{Q}{T}

        \int dS = \int \frac{Q}{T}

Also we know that at constant pressure, Q = \Delta H = C_{p} - dT

     \int_{S_{1}}^{S_{2}} dS = \int_{T_{1}}^{T_{2}} C_{p} \frac{dT}{T}

     \Delta S = C_{p} \int_{T_{1}}^{T_{2}} \frac{dT}{T}

As the given data is as follows.

        T_{1} = 298 K,          T_{2} = 348 K

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Now, putting the given values into the above formula as follows.

          \Delta S = C_{p} ln {T_{1}}^{T_{2}} \frac{dT}{T}

                  = 29.355 [ln (348) - ln (298)]

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Thus, we can conclude that the increase in the molar entropy of given oxygen gas is 4.48 J/k mol.

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Answer:

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4 0
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Explanation :

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ivann1987 [24]

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