200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralization of this reaction is -57.1 kJ. The increase in temperature in °C of the system on mixing is x > 10^-2. The value of x is__________. (Nearest integer)

[Given : Specific heat of water = 4.18 J g^-1 K^-1

Density of water = 1.00 g cm^-3]

(Assume no volume change on mixing)

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Vishal Baghel | Contributor-Level 10

2 months ago

Millimoles of HCl = 200 × 0.2 = 40

Millimoles of NaOH = 300 × 0.1 = 30

Heat released = $\frac{30}{1000} \times 57.1 \times 1000 J = 1713 J$

$[\begin{array}{l} ρ = \frac{m}{v} \\ m = ρ v \end{array}]$

Mass of solution = 500 × 1 g

= 500 g

Specific heat of water = 4.18 Jg^-1 K^-1

$Δ T = \frac{q}{m c}$

$= \frac{1713}{500 \times 4.18} ° C$

$= 81.96 \times 1 0^{- 2} ° C \approx 82 \times 1 0^{- 2} ° C$

Ans. = 82

Millimoles of HCl = 200 × 0.2 = 40Millimoles of NaOH = 300 × 0.1 = 30Heat released =  <math> <mrow> <mfrac> <mrow> <mn>3</mn> <mn>0</mn> </mrow> <mrow> <mn>1</mn> <mn>0</mn> <mn>0</mn> <mn>0</mn> </mrow> </mfrac> <mo>×</mo> <mn>5</mn> <mn>7</mn> <mo>.</mo> <mn>1</mn> <mo>×</mo> <mn>1</mn> <mn>0</mn> <mn>0</mn> <mn>0</mn> <mi>J</mi> <mo>=</mo> <mn>1</mn> <mn>7</mn> <mn>1</mn> <mn>3</mn> <mi>J</mi> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1815915462"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mrow> <mo> [</mo> <mtable columnalign="left"> <mtr> <mtd> <mrow> <mi>ρ</mi> <mo>=</mo> <mfrac> <mrow> <mi>m</mi> </mrow> <mrow> <mi>v</mi> </mrow> </mfrac> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mi>m</mi> <mo>=</mo> <mi>ρ</mi> <mi>v</mi> </mrow> </mtd> </mtr> </mtable> <mo>]</mo> </mrow> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1815915463"> </o:OLEObject></xml><! [endif]->Mass of solution = 500 × 1 g= 500 gSpecific heat of water = 4.18 Jg-1 K-1 <math> <mrow> <mi>Δ</mi> <mi>T</mi> <mo>=</mo> <mfrac> <mrow> <mi>q</mi> </mrow> <mrow> <mi>m</mi> <mi>c</mi> </mrow> </mfrac> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1815915464"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mn>7</mn> <mn>1</mn> <mn>3</mn> </mrow> <mrow> <mn>5</mn> <mn>0</mn> <mn>0</mn> <mo>×</mo> <mn>4</mn> <mo>.</mo> <mn>1</mn> <mn>8</mn> </mrow> </mfrac> <mo>°</mo> <mi>C</mi> </mrow> </math>  <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1028" DrawAspect="Content" ObjectID="_1815915465"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mo>=</mo> <mn>8</mn> <mn>1</mn> <mo>.</mo> <mn>9</mn> <mn>6</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>2</mn> </mrow> </msup> <mo>°</mo> <mi>C</mi> <mo>≈</mo> <mn>8</mn> <mn>2</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>2</mn> </mrow> </msup> <mo>°</mo> <mi>C</mi> </mrow> </math> Ans. = 82

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