Question

The molarity of Lit of a solution that contains 65.5 ppm of Li2CO3 MW = 73.9 g / mol Select one: O a. 5.65X10-3 O b. 1.77X10-3 O c. 8.86X10-4 O d. 9.05X10-4 O e. 7.39X10-3

The following results were obtained in the replicate analysis of lead content in water: 0.752, 0.756, 0.752, 0.751, and 0.760 ppm. If the standard deviation is 0.004 ppm, the confidence limits at 95% level is t = 2.02 at degree of freedom 5, and 2.13 at degree of freedom 4 Select one: O a. 0.754 + 1.6x10-3 O b. 0.754 + 5.2x10-2 O c. 0.754 3.8x10-3. d. 0.754 4.2x10-3 O e. 0.754 + 5.0 * 10-3

An analysis of a sample containing silver has an error of 0.4 mg higher than the actual value which is 500 mg the percent relative error is Select one: O a. 0.4% O b. 0.08% O C. -0.04% O d. 1% O e. -0.08%

Answer 1

As 1 ppm is equal to the 1 microgram /ml so 65.5 ppm solution is having the 65.5 microgram /ml

which is equal to 65.5 mg/L

now 73.9 g/ mole is the molecular weight of the Li2CO3 so 65.5 mg contains = 65.5 /73.9 = 0.88633 mili mole.

which is equal to 8.86 *10^(-4) moles

So 8.86 *10^(-4) moles contains in the 1000 ml so molarity is 8.86 *10^(-4) M

So correct answer is Option C 8.86 *10^(-4) M

Answer 2:

If we calculate the mean :

(0.752+0.754+0.752+0.751+0.760)/5 = 0.7542 ppm

Which is equal to 0.754 ppm

Now as we have 5 observation so standard error = standard deviation /sqrt(smaple size) = 0.004 /sqrt(5) = 0.001789

As we have 5 sample size so we need to have four degree of the freedom so we need to choos the t value is 2.13

Now the 95% confidence interaval is

lower limit = mean - standard error *t value = 0.754- 2.13 *0.001789 = 0.754-0.0038

upper limit = mean + standard error *t value= 0.754+ 2.13 *0.001789 = 0.754+0.0038

So correct answer is 0.754 ,3.8 *10^(-3)

So correct answer is option C :0.754 ,3.8 *10^(-3)

Answer 3:

Percent relative error = |actual value -expected value|*100/|Expected value

= |500.4 -500|*100/500 = 0.4/5 = 0.08 %

So correct answer is option B 0.08 %