Ihave a 5 m stock solution of kci (for a protocol 100 mm kcl is considered "1x" concentration). if i want to make 10 ml of a 4x concentration kcl solution, how much 5 m stock and how much water do i need to add together?

You need 0.8 ml of 5M stock solution and you have to add 9.2 ml of water.

Explanation:

Protocol solution (1X): 100 mM=0.1M

4X: 0.4M

The concentration of a solution is inversely proportional to the volume of a solution, so:

where:

M1= 5M stock solution

V1= amount of solution we need to collect

M2=4X solution

V2= 10 ml (volumen of 4X solution)

Therefore:

5M×V1=0.4M×10ml

V1={0.4M}{5M}10ml=0.8ml

To make a 10 ml solution we have to add 9.2 ml of water because V2 es 10 ml.

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explanation:

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