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siniylev [52]
3 years ago
7

For a titration of 25 ml of naoh solution, 23.30 ml of hcl solution if used. what is the molarity of naoh

Chemistry
1 answer:
Eduardwww [97]3 years ago
3 0

<span>HCl<span>(aq)</span>+NaOH<span>(aq)</span>→NaCl<span>(aq)</span>+<span>H2</span>O<span>(l)</span></span>

As you can see here, one mole of acid neutralizes one mole of base.

We use the concentration equation, which states that,

<span>c=<span>nv</span></span>

<span> <span> <span> n is the number of moles </span> <span> v is the volume of solution </span> </span> </span>

Rearranging for moles, we get,

<span>n=c⋅v</span>

So, we have:

<span><span>n<span>NaOH</span></span>=0.1 M⋅0.05 L</span>

<span>=0.005 mol</span>

Since one mole of acid neutralizes one mole of base, then we must have: <span><span>n<span>HCl</span></span>=<span>n<span>NaOH</span></span></span>.

And so,

<span><span>c<span>HCl</span></span>=<span><span>n<span>HCl</span></span><span>v<span>HCl</span></span></span></span>

<span>=<span><span>0.005 mol</span><span>0.03 L</span></span></span>

<span>≈0.17 <span>M</span></span>

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In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

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