A red sports drink contains Red 40 dye. A 5.4 mL aliquot of this sports drink was diluted to 25.0 mL with deionized water in a volumetric flask to form a dilute drink solution. The concentration of Red 40 dye in the dilute solution was determined to be 18.1 ppm from the calibration plot. What is the concentration of Red 40 dye (ppm) in the original sports drink

84.0 ppm is the concentration of Red 40 dy in the original sports drink.

Explanation:

Concentration of red dye in sport drink before dilution

Volume of the sport drink before dilution

Concentration of red dye in sport drink after dilution

Volume of the sport drink after dilution

( dilution )

84.0 ppm is the concentration of Red 40 dy in the original sports drink.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

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explanation:

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